（并查集状态压缩）CodeForces

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.

注意到如果x存在于A果a-x、b-x（如果存在）也必须存在在A，如果a-(b-x) 、b-(a-x)存在也必须在A……即所有数可以划分为若干个集合，每个集合中的元素必须同时属于A或B。这可以通过并查集维护。如果某数x，a-x,b-x都不存在那么这个数无法放到A、B任何一个，直接输出NO即可。

接下来维护并查集的状态，采用状态压缩，用两位2进制数表示可否放到A、B集合中。对于每个数x，如果a-x存在则其可以放在集合A，如果b-x存在则其可以放在集合B。对每个并查集取其集合内所有可行状态的交集。显然并查集之间是不再存在互相影响的。如果每个并查集可行状态非0就表明一定存在符合题意的分配方式，任取每个集合的任意可行分配即可。

  1 #include <iostream>
  2 #include <string>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <queue>
  8 #include <set>
  9 #include <map>
 10 #include <list>
 11 #include <vector>
 12 #include <stack>
 13 #define mp make_pair
 14 //#define P make_pair
 15 #define MIN(a,b) (a>b?b:a)
 16 //#define MAX(a,b) (a>b?a:b)
 17 typedef long long ll;
 18 typedef unsigned long long ull;
 19 const int MAX=1e5+5;
 20 const int MAX_V=25;
 21 const int INF=2e9+5;
 22 const double M=4e18;
 23 using namespace std;
 24 const int MOD=1e9+7;
 25 typedef pair<ll,int> pii;
 26 const double eps=0.000000001;
 27 #define rank rankk
 28 map<int,int> par;//父亲
 29 map<int,int> rank;//树的高度
 30 //初始化n个元素
 31 map<int,bool> chu;
 32 map<int,int> st;
 33 int an[MAX];
 34 void init(int n)
 35 {
 36     for(int i=1;i<n;i++)
 37     {
 38         par[i]=i;
 39         rank[i]=0;
 40     }
 41 }
 42 //查询树的根，期间加入了路径压缩
 43 int find(int x)
 44 {
 45     if(!par[x])
 46     {
 47         rank[x]=0;
 48         return par[x]=x;
 49     }
 50     if(par[x]==x)
 51         return x;
 52     else
 53         return par[x]=find(par[x]);
 54 }
 55 //合并x和y所属的集合
 56 void unite(int x,int y)
 57 {
 58     x=find(x);
 59     y=find(y);
 60     if(x==y)
 61         return ;
 62     if(rank[x]<rank[y])
 63         par[x]=y;
 64     else
 65     {
 66         par[y]=x;
 67         if(rank[x]==rank[y])
 68             rank[x]++;
 69     }
 70 }
 71 //判断x和y是否属于同一个集合
 72 bool same(int x,int y)
 73 {
 74     return find(x)==find(y);
 75 }
 76 int n,a,b;
 77 int x[MAX];
 78 int main()
 79 {
 80     scanf("%d%d%d",&n,&a,&b);
 81     for(int i=1;i<=n;i++)
 82     {
 83         scanf("%d",&x[i]);
 84         chu[x[i]]=true;
 85     }
 86     for(int i=1;i<=n;i++)
 87         st[x[i]]=0;
 88     for(int i=1;i<=n;i++)
 89     {
 90         bool sta=false;
 91         if(chu[a-x[i]])
 92         {
 93             sta=true;
 94             unite(x[i],a-x[i]);
 95             st[x[i]]|=1;
 96         }
 97         if(chu[b-x[i]])
 98         {
 99             sta=true;
100             unite(x[i],b-x[i]);
101             st[x[i]]|=2;
102         }
103         if(!sta)
104             return 0*printf("NO
");
105     }
106     for(int i=1;i<=n;i++)
107     {
108         st[find(x[i])]&=st[x[i]];
109     }
110     for(int i=1;i<=n;i++)
111         if(!st[x[i]])
112             return 0*printf("NO
");
113     printf("YES
");
114     for(int i=1;i<=n;i++)
115     {
116         if(st[find(x[i])]&1)
117             printf("0 ");
118         else
119             printf("1 ");
120     }
121     printf("
");
122     return 0;
123 }