/*
2SAT，找出一组解
同3648
注意一点的是，在最后输入答案的时候犯了很多错误。
最后只输出颜色为1的点
如果没有目标点的话，就在下行找颜色为1的点
像3648,需要找和新娘颜色一样的点，新娘的颜色一定为1，但是其在上行，所以在上行中找颜色为1的点
*/
// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>
#include <strstream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>
using namespace std;
// typedef
typedef long long LL;
typedef unsigned long long ULL;
//
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FF(i,a) for(int i=0;i<(a);i+++)
#define FFD(i,a) for(int i=(a)-1;i>=0;i--)
#define Z(a) (a<<1)
#define Y(a) (a>>1)
const double eps = 1e-11;
const double Pi = acos(-1.0);
template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
if(x>y) return x;
return y;
}
template<class T> inline T TMIN(T x,T y)
{
if(x<y) return x;
return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
T t = x;
x = y;
y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
return TMAX(TMAX(x,y),z);
}
// code begin
#define MAXN 5010
struct node
{
int s;
int e;
int number;
};
node tm[MAXN];
vector<int> G[MAXN];
vector<int> NG[MAXN];
vector<int> tp;
int in[MAXN];
int col[MAXN];
int link[MAXN];
int scc;
int cnt;
int used[MAXN];
int stk1[MAXN],top1;
int stk2[MAXN],top2;
int isin[MAXN];
int dfn[MAXN];
int low[MAXN];
int id[MAXN];
int N;
void gabow_scc(int i)
{
used[i] = true;
stk1[top1++] = i;
stk2[top2++] = i;
dfn[i] = cnt++;
isin[i] = true;
FORj( 0,G[i].size(),1 )
{
if(!used[ G[i][j] ])
{
gabow_scc(G[i][j]);
}
else if(isin[G[i][j]])
{
while(dfn[stk2[top2-1]]>dfn[G[i][j]])
top2--;
}
}
if(i==stk2[top2-1])
{
top2--;
int w;
do
{
w = stk1[--top1];
isin[w] = false;
id[w] = scc;
}while(w!=i);
scc++;
}
}
void tarjan_scc(int i)
{
used[i] = true;
stk1[top1++] = i;
dfn[i] = cnt;
low[i] = cnt;
cnt++;
isin[i] = true;
FORj(0,G[i].size(),1)
{
if(!used[ G[i][j] ])
{
tarjan_scc(G[i][j]);
low[i] = TMIN(low[i],low[G[i][j]]);
}
else if(isin[G[i][j]])
{
low[i] = TMIN(low[i],dfn[G[i][j]]);
}
}
if(dfn[i]==low[i])
{
int w;
do
{
w=stk1[--top1];
isin[w] = false;
id[w] = scc;
}while(w!=i);
scc++;
}
}
bool Isok(int a,int b)
{
int sa = tm[a].s;
int sb = tm[b].s;
int ea = tm[a].e;
int eb = tm[b].e;
if( sa<eb&&sa>=sb ) return true;
if( ea<=eb && ea>sb ) return true;
if( sb<ea&&sb>=sa ) return true;
if( eb<=ea && eb>sa ) return true;
return false;
}
void dfs(int i)
{
FORj(0,NG[i].size(),1)
{
if(!col[NG[i][j]])
{
col[NG[i][j]] = 2;
dfs(NG[i][j]);
}
}
}
int main()
{
read;
write;
int sh,sm,eh,em,po;
while(scanf("%d",&N)!=-1)
{
int j=0;
FORi(0,N,1)
{
scanf("%d:%d %d:%d %d",&sh,&sm,&eh,&em,&po);
tm[j].s = sh*60+sm;
tm[j].e = tm[j].s+po;
tm[j].number = i;
j++;
tm[j].e = eh*60+em;
tm[j].s = tm[j].e-po;
tm[j].number = i;
j++;
}
/*
FORi(0,2*N,1)
{
printf("%02d:%02d %02d:%02d\n",tm[i].sh,tm[i].sm,tm[i].eh,tm[i].em);
}
printf("\n");
*/
//建图
FORi(0,4*N,1)
{
G[i].clear();
}
//判断时间区间的限制性条件
FORi(0,2*N,2)
{
FORj(i+2,2*N,1)
{
// i ,j
if( Isok(i,j) )
{
//printf("相交");
//printf(" %02d:%02d %02d:%02d",tm[i].sh,tm[i].sm,tm[i].eh,tm[i].em);
//printf(" %02d:%02d %02d:%02d\n",tm[j].sh,tm[j].sm,tm[j].eh,tm[j].em);
G[2*i+1].push_back(2*j+0);
G[2*j+1].push_back(2*i+0);
}
}
FORj(i+2,2*N,1)
{
// i ,j
if( Isok(i+1,j) )
{
//printf("相交");
//printf(" %02d:%02d %02d:%02d",tm[i].sh,tm[i].sm,tm[i].eh,tm[i].em);
//printf(" %02d:%02d %02d:%02d\n",tm[j].sh,tm[j].sm,tm[j].eh,tm[j].em);
G[2*(i+1)+1].push_back(2*j+0);
G[2*j+1].push_back(2*(i+1)+0);
}
}
}
FORi(0,2*N,2)
{
//i i+1
G[2*i].push_back(2*(i+1)+1);
G[2*(i+1)+1].push_back(2*i);
G[2*i+1].push_back(2*(i+1));
G[2*(i+1)].push_back(2*i+1);
}
// scc
memset(used,0,sizeof(used));
memset(id,0,sizeof(id));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(isin,0,sizeof(isin));
top1 = 0;
scc = 1;
cnt = 1;
FORi(0,4*N,1)
{
if(!used[i])
{
tarjan_scc(i);
}
}
// 判断
bool f= true;
FORi(0,2*N,1)
{
if(id[2*i]==id[2*i+1])
{
f=false;
break;
}
}
if(f)
{
//找一组解
printf("YES\n");
FORi(1,scc,1)
{
NG[i].clear();
}
//缩图
memset(link,0,sizeof(link));
FORi(0,4*N,1)
{
FORj(0,G[i].size(),1)
{
if( id[i]!=id[G[i][j]] )
{
NG[id[G[i][j]]].push_back(id[i]);
}
}
}
//对立面
FORi(0,2*N,1)
{
link[ id[2*i] ] = id[2*i+1];
link[ id[2*i+1] ] = id[2*i];
}
//取出重复
memset(in,0,sizeof(in));
FORi(1,scc,1)
{
vector<int> tmp = NG[i];
NG[i].clear();
if(tmp.size()>0)
{
sort(tmp.begin(),tmp.end());
NG[i].push_back(tmp[0]);
in[tmp[0]]++;
FORj(1,tmp.size(),1)
{
if(tmp[j]!=tmp[j-1])
{
NG[i].push_back(tmp[j]);
in[tmp[j]]++;
}
}
}
}
//新图NG
tp.clear();
queue<int> qi;
FORi(1,scc,1)
{
if(in[i]==0)
qi.push(i);
}
while(!qi.empty())
{
int cur = qi.front();
qi.pop();
tp.push_back(cur);
FORi(0,NG[cur].size(),1)
{
in[ NG[cur][i] ]--;
if( in[ NG[cur][i] ]==0 )
{
qi.push( NG[cur][i] );
}
}
}
// 根据拓扑序，找解
memset(col,0,sizeof(col));
//printf("shang\n");
//printf("tp.size:%d scc:%d\n",tp.size(),scc);
FORi(0,tp.size(),1)
{
if(!col[tp[i]])
{
col[tp[i]] = 1;
//
col[link[tp[i]]] = 2;
//
dfs(link[tp[i]]);
}
}
FORi(0,2*N,2)
{
//printf("col %d: %d %d %02d:%02d %02d:%02d\n",i,col[id[2*i+0]],col[id[2*i+1]],tm[i].s/60,tm[i].s%60,tm[i].e/60,tm[i].e%60);
//printf("col %d: %d %d %02d:%02d %02d:%02d\n",i+1,col[id[2*(i+1)+0]],col[id[2*(i+1)+1]],tm[i+1].s/60,tm[i+1].s%60,tm[i+1].e/60,tm[i+1].e%60);
//printf("col:%d %d\n",col[id[2*i+1]],col[id[2*0+1]]);
if( col[id[2*i+1]]==1 ) //col[id[2*0+1]]
printf("%02d:%02d %02d:%02d\n",tm[i].s/60,tm[i].s%60,tm[i].e/60,tm[i].e%60);
//printf("col:%d %d\n",col[id[2*(i+1)+1]],col[id[2*0+1]]);
if( col[id[2*(i+1)+1]]==1 )
printf("%02d:%02d %02d:%02d\n",tm[i+1].s/60,tm[i+1].s%60,tm[i+1].e/60,tm[i+1].e%60);
}
}
else printf("NO\n");
//printf("\n");
}
return 0;
}
