Two Sets CodeForces - 468B

Little X has n distinct integers: p₁, p₂, ..., p_n. He wants to divide all of them into two sets A and B. The following two conditions must be satisfied:

• If number x belongs to set A, then number a - x must also belong to set A.
• If number x belongs to set B, then number b - x must also belong to set B.

Help Little X divide the numbers into two sets or determine that it's impossible.

Input

The first line contains three space-separated integers n, a, b (1 ≤ n ≤ 10⁵; 1 ≤ a, b ≤ 10⁹). The next line contains n space-separated distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ 10⁹).

Output

If there is a way to divide the numbers into two sets, then print "YES" in the first line. Then print n integers: b₁, b₂, ..., b_n (b_i equals either 0, or 1), describing the division. If b_i equals to 0, then p_i belongs to set A, otherwise it belongs to set B.

If it's impossible, print "NO" (without the quotes).

Example

Input

4 5 9
2 3 4 5

Output

YES
0 0 1 1

Input

3 3 4
1 2 4

Output

NO

Note

It's OK if all the numbers are in the same set, and the other one is empty.

增加一个案列：

4  10  16

4   6   10   12

题解：哇，被坑惨了。用二分搜索调了好久才发现有些情况解决不了，比如a-x和b-x同时存在时，不能直接确定放在那个位置。后面看了

别人的题解（http://www.cnblogs.com/zhien-aa/p/6726862.html），才知道用并查集管理。

#include<bits/stdc++.h>
using namespace std;

const int maxn=1e5+5; 

int n,a,b;
int F[maxn],num[maxn];

int Find(int x){
    if(x==F[x]) return F[x];
    return F[x]=Find(F[x]);
}

int main()
{   while(cin>>n>>a>>b){
        map<int,int> p;
        for(int i=1;i<=n+2;i++) F[i]=i;
        for(int i=1;i<=n;i++){
            cin>>num[i];
            p[num[i]]=i;
        }
        int temp;
        for(int i=1;i<=n;i++){
            temp=Find(i);
            if(p[a-num[i]]) F[temp]=Find(p[a-num[i]]);
            else F[temp]=Find(n+1);
            temp=Find(i);
            if(p[b-num[i]]) F[temp]=Find(p[b-num[i]]);
            else F[temp]=Find(n+2);
        }
        if(Find(n+1)==Find(n+2)) cout<<"NO"<<endl;
        else{
            cout<<"YES"<<endl;
            for(int i=1;i<=n;i++) printf("%d%c",Find(i)==Find(n+1),i==n?'\n':' '); 
        }
    }
}