CF982C Cut 'em all! DFS 树 * 二十一

 Cut 'em all!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given a tree with $\mathrm{nn vertices.}$

Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.

Input

The first line contains an integer $\mathrm{nn (1\le n\le 1051\le n\le 105)\; denoting\; the\; size\; of\; the\; tree.}$

The next $\mathrm{n-1n-1 lines\; contain\; two\; integers uu, vv (1\le u,v\le n1\le u,v\le n)\; each,\; describing\; the\; vertices\; connected\; by\; the ii-th\; edge.}$

It's guaranteed that the given edges form a tree.

Output

Output a single integer $\mathrm{kk —\; the\; maximum\; number\; of\; edges\; that\; can\; be\; removed\; to\; leave\; all\; connected\; components\; with\; even\; size,\; or -1-1 if\; it\; is\; impossible\; to\; remove\; edges\; in\; order\; to\; satisfy\; this\; property.}$

Examples

input

Copy

4
2 4
4 1
3 1

output

Copy

1

input

Copy

3
1 2
1 3

output

Copy

-1

input

Copy

10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5

output

Copy

4

input

Copy

2
1 2

output

Copy

0

Note

In the first example you can remove the edge between vertices $\mathrm{11 and 44.\; The\; graph\; after\; that\; will\; have\; two\; connected\; components\; with\; two\; vertices\; in\; each.}$

In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is $-1-1.$

 

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int hd[maxn], ne[maxn*2], to[maxn*2], num, n, siz[maxn], ans;
void add( int x, int y ) {
    to[++num]=y, ne[num]=hd[x], hd[x]=num;
}
 
void dfs( int x, int fa ) {
    siz[x]=1;
    for( int i = hd[x]; i; i = ne[i] ) {
        if(to[i]!=fa){
            dfs(to[i],x);
            siz[x]+=siz[to[i]];
        }
    }
    if(!(siz[x]&1)) 
        siz[x]=0,ans++;
}
 
int main(){
    std::ios::sync_with_stdio(false);
    scanf("%d",&n);
    int uu, vv;
    for( int i = 1; i < n; i ++ ) {
        scanf("%d%d",&uu,&vv);
        add(uu,vv), add(vv,uu);
    }
    if( n & 1 ) { 
        puts("-1"); 
        return 0;
    }
     
    dfs( 1, -1 ), ans--;
     
    printf("%d\n", ans );
    return 0;
}