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

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

Example

4 5 9
2 3 4 5

YES
0 0 1 1

3 3 4
1 2 4

NO

Note

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

增加一個案列：

4 10 16

4 6 10 12

題解：哇，被坑慘了。用二分搜索調了好久才發現有些情況解決不了，比如a-x和b-x同時存在時，不能直接確定放在那個位置。後面看了

別人的題解（http://www.cnblogs.com/zhien-aa/p/6726862.html），才知道用並查集管理。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn=1e5+5; 
 5 
 6 int n,a,b;
 7 int F[maxn],num[maxn];
 8 
 9 int Find(int x){
10     if(x==F[x]) return F[x];
11     return F[x]=Find(F[x]);
12 }
13 
14 int main()
15 {   while(cin>>n>>a>>b){
16         map<int,int> p;
17         for(int i=1;i<=n+2;i++) F[i]=i;
18         for(int i=1;i<=n;i++){
19             cin>>num[i];
20             p[num[i]]=i;
21         }
22         int temp;
23         for(int i=1;i<=n;i++){
24             temp=Find(i);
25             if(p[a-num[i]]) F[temp]=Find(p[a-num[i]]);
26             else F[temp]=Find(n+1);
27             temp=Find(i);
28             if(p[b-num[i]]) F[temp]=Find(p[b-num[i]]);
29             else F[temp]=Find(n+2);
30         }
31         if(Find(n+1)==Find(n+2)) cout<<"NO"<<endl;
32         else{
33             cout<<"YES"<<endl;
34             for(int i=1;i<=n;i++) printf("%d%c",Find(i)==Find(n+1),i==n?‘\n‘:‘ ‘); 
35         }
36     }
37 } 

Two Sets CodeForces - 468B