cf468B Two Sets

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).

Examples

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.

如果A中有了一个x，那么A中也要有a-x，说明（逆否命题）如果A中没有a-x，也就没有x。即如果B中有a-x，就有x。因为不在A就在B咯

所以这个A还是B无所谓的，重要的是x和a-x一定在同一集合，x和b-x一定在同一集合

因此裸并查集，只要被并起来的数字有一个不能在A，那么这一群都不能在A

#include<bits/stdc++.h>
#define LL long long
using namespace std;
LL n,a,b;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct po{LL x;int rnk;}p[100010];
bool operator <(po a,po b){return a.x<b.x;}
int fa[100010];
int mrk[100010];
inline int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
map<LL,LL>mp;
int main()
{
    n=read();a=read();b=read();
    for (int i=1;i<=n;i++)
    {
        p[i].x=read();
        p[i].rnk=i;
    }
    sort(p+1,p+n+1);
    for(int i=1;i<=n;i++)mp[p[i].x]=p[i].rnk,fa[i]=i,mrk[i]=3;
    for(int i=1;i<=n;i++)
    {
        if (mp[a-p[i].x])
        {
            int pos=getfa(mp[a-p[i].x]),pos2=getfa(p[i].rnk);
            if (pos!=pos2)fa[pos2]=pos;
        }
        if (mp[b-p[i].x])
        {
            int pos=getfa(mp[b-p[i].x]),pos2=getfa(p[i].rnk);
            if (pos!=pos2)fa[pos2]=pos;
        }
    }
    for (int i=1;i<=n;i++)
    {
        int ff=getfa(p[i].rnk);
        if (!mp[a-p[i].x])mrk[ff]&=1;
        if (!mp[b-p[i].x])mrk[ff]&=2;
    }
    for (int i=1;i<=n;i++)
        if (mrk[getfa(i)]==0){puts("NO");return 0;}
        else if (mrk[getfa(i)]==3)mrk[getfa(i)]=1;
    puts("YES");
    for (int i=1;i<=n;i++)printf("%d ",mrk[getfa(i)]==2?0:1);
    puts("");
}