#include <bits/stdc++.h>
using namespace std;
int N, M;
int n1, n2, n3;
int C[2], no[2];
vector<pair<int, int> > P, F;
vector<int> V[5003];
int mark[5003], dp[5003][5003];
bool dfs(int nod, int pad, int color) {
  mark[nod] = color + 1;
  if (no[color] == 0) no[color] = nod;
  C[color]++;
  for (auto &v : V[nod]) {
    if (v == pad) continue;
    if (mark[nod] == mark[v]) return false;
    if (mark[v]) continue;
    if (!dfs(v, nod, (color + 1) % 2)) return false;
  }
  return true;
}
void marking(int nod, int pad, int color) {
  if (color == 0)
    mark[nod] = 2;
  else {
    if (n3 > 0)
      mark[nod] = 3, n3--;
    else
      mark[nod] = 1;
  }
  for (auto &v : V[nod]) {
    if (v == pad) continue;
    if (mark[v]) continue;
    marking(v, nod, (color + 1) % 2);
  }
}
bool findPairs() {
  int a, b;
  for (int i = 0; i < M; i++) {
    cin >> a >> b;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (!mark[i]) {
      C[0] = C[1] = 0;
      no[0] = no[1] = 0;
      if (!dfs(i, 0, 0)) return false;
      P.push_back(pair<int, int>(C[0], C[1]));
      F.push_back(pair<int, int>(no[0], no[1]));
    }
  for (int i = 0; i <= N; i++) mark[i] = 0;
  dp[0][0] = 1;
  a = 0;
  for (auto &p : P) {
    for (int i = n2; i >= 0; i--) {
      if (i - p.first >= 0 && dp[a][i - p.first]) dp[a + 1][i] = 1;
      if (i - p.second >= 0 && dp[a][i - p.second]) dp[a + 1][i] = 2;
    }
    a++;
  }
  if (dp[a][n2] == 0) return false;
  b = n2;
  for (int i = a; i >= 1; i--) {
    if (dp[i][b] == 1) {
      marking(F[i - 1].first, 0, 0);
      b -= P[i - 1].first;
    } else {
      marking(F[i - 1].second, 0, 0);
      b -= P[i - 1].second;
    }
  }
  for (int i = 1; i <= N; i++)
    if (!mark[i]) {
      if (n3 > 0)
        mark[i] = 3, n3--;
      else
        mark[i] = 1;
    }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M >> n1 >> n2 >> n3;
  if (!findPairs())
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= N; i++) cout << mark[i];
    cout << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

You are given an undirected graph without self-loops or multiple edges which consists of n vertices and m edges. Also you are given three integers n_1, n_2 and n_3.

Can you label each vertex with one of three numbers 1, 2 or 3 in such way, that: 

  1. Each vertex should be labeled by exactly one number 1, 2 or 3; 
  2. The total number of vertices with label 1 should be equal to n_1; 
  3. The total number of vertices with label 2 should be equal to n_2; 
  4. The total number of vertices with label 3 should be equal to n_3; 
  5. |col_u - col_v| = 1 for each edge (u, v), where col_x is the label of vertex x. 



If there are multiple valid labelings, print any of them.

Input

The first line contains two integers n and m (1 ≤ n ≤ 5000; 0 ≤ m ≤ 10^5) — the number of vertices and edges in the graph.

The second line contains three integers n_1, n_2 and n_3 (0 ≤ n_1, n_2, n_3 ≤ n) — the number of labels 1, 2 and 3, respectively. It's guaranteed that n_1 + n_2 + n_3 = n.

Next m lines contan description of edges: the i-th line contains two integers u_i, v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) — the vertices the i-th edge connects. It's guaranteed that the graph doesn't contain self-loops or multiple edges.

Output

If valid labeling exists then print "YES" (without quotes) in the first line. In the second line print string of length n consisting of 1, 2 and 3. The i-th letter should be equal to the label of the i-th vertex.

If there is no valid labeling, print "NO" (without quotes).

Examples

Input


6 3
2 2 2
3 1
5 4
2 5


Output


YES
112323


Input


5 9
0 2 3
1 2
1 3
1 5
2 3
2 4
2 5
3 4
3 5
4 5


Output


NO

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int N, M;
int n1, n2, n3;
int C[2], no[2];
vector<pair<int, int> > P, F;
vector<int> V[5003];
int mark[5003], dp[5003][5003];
bool dfs(int nod, int pad, int color) {
  mark[nod] = color + 1;
  if (no[color] == 0) no[color] = nod;
  C[color]++;
  for (auto &v : V[nod]) {
    if (v == pad) continue;
    if (mark[nod] == mark[v]) return false;
    if (mark[v]) continue;
    if (!dfs(v, nod, (color + 1) % 2)) return false;
  }
  return true;
}
void marking(int nod, int pad, int color) {
  if (color == 0)
    mark[nod] = 2;
  else {
    if (n3 > 0)
      mark[nod] = 3, n3--;
    else
      mark[nod] = 1;
  }
  for (auto &v : V[nod]) {
    if (v == pad) continue;
    if (mark[v]) continue;
    marking(v, nod, (color + 1) % 2);
  }
}
bool findPairs() {
  int a, b;
  for (int i = 0; i < M; i++) {
    cin >> a >> b;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  for (int i = 1; i <= N; i++)
    if (!mark[i]) {
      C[0] = C[1] = 0;
      no[0] = no[1] = 0;
      if (!dfs(i, 0, 0)) return false;
      P.push_back(pair<int, int>(C[0], C[1]));
      F.push_back(pair<int, int>(no[0], no[1]));
    }
  for (int i = 0; i <= N; i++) mark[i] = 0;
  dp[0][0] = 1;
  a = 0;
  for (auto &p : P) {
    for (int i = n2; i >= 0; i--) {
      if (i - p.first >= 0 && dp[a][i - p.first]) dp[a + 1][i] = 1;
      if (i - p.second >= 0 && dp[a][i - p.second]) dp[a + 1][i] = 2;
    }
    a++;
  }
  if (dp[a][n2] == 0) return false;
  b = n2;
  for (int i = a; i >= 1; i--) {
    if (dp[i][b] == 1) {
      marking(F[i - 1].first, 0, 0);
      b -= P[i - 1].first;
    } else {
      marking(F[i - 1].second, 0, 0);
      b -= P[i - 1].second;
    }
  }
  for (int i = 1; i <= N; i++)
    if (!mark[i]) {
      if (n3 > 0)
        mark[i] = 3, n3--;
      else
        mark[i] = 1;
    }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M >> n1 >> n2 >> n3;
  if (!findPairs())
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= N; i++) cout << mark[i];
    cout << "\n";
  }
  return 0;
}

```