#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const int mm = 5005;
vector<int> graph[mm], con[mm], st, color(mm, -1), vis(mm, 0);
int n, m, a, b, c, cnt[mm][2];
int x, y;
bool no = 0;
void dfs(int node, int col, int root) {
  con[root].push_back(node);
  vis[node] = 1;
  color[node] = col;
  cnt[root][col]++;
  for (int i : graph[node]) {
    if (color[i] == color[node]) {
      no = 1;
      return;
    }
    if (vis[i] == 0) dfs(i, ((col == 0) ? 1 : 0), root);
  }
}
int memo[mm][mm], dir[mm][mm];
int dp(int ind, int sum) {
  if (sum < 0) return 0;
  if (ind == st.size()) {
    if (sum == 0)
      return 1;
    else
      return 0;
  }
  if (memo[ind][sum] != -1) return memo[ind][sum];
  int node = st[ind];
  int c1 = dp(ind + 1, sum - cnt[node][0]);
  int c2 = dp(ind + 1, sum - cnt[node][1]);
  if (c1 != 0) dir[ind][sum] = 0;
  if (c2 != 0) dir[ind][sum] = 1;
  return memo[ind][sum] = (c1 || c2);
}
int main() {
  cin >> n >> m;
  cin >> a >> b >> c;
  memset(cnt, 0, sizeof(cnt));
  memset(memo, -1, sizeof(memo));
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  x = a + c;
  y = b;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      st.push_back(i);
      dfs(i, 0, i);
    }
  }
  if (no or !dp(0, b)) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    int sum = b;
    for (int i = 0; i < st.size(); i++) {
      for (int j : con[st[i]]) {
        if (color[j] == dir[i][sum] and b > 0) {
          color[j] = 2;
          --b;
        } else {
          if (a > 0) {
            color[j] = 1;
            --a;
          } else {
            color[j] = 3;
            --c;
          }
        }
      }
      sum -= cnt[st[i]][dir[i][sum]];
    }
    if (a != 0) cout << a << endl;
    for (int i = 1; i <= n; i++) cout << color[i];
    cout << endl;
  }
}

### Prompt

Your task is to create a Cpp solution to 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;
using ll = long long;
using ull = unsigned long long;
const int mm = 5005;
vector<int> graph[mm], con[mm], st, color(mm, -1), vis(mm, 0);
int n, m, a, b, c, cnt[mm][2];
int x, y;
bool no = 0;
void dfs(int node, int col, int root) {
  con[root].push_back(node);
  vis[node] = 1;
  color[node] = col;
  cnt[root][col]++;
  for (int i : graph[node]) {
    if (color[i] == color[node]) {
      no = 1;
      return;
    }
    if (vis[i] == 0) dfs(i, ((col == 0) ? 1 : 0), root);
  }
}
int memo[mm][mm], dir[mm][mm];
int dp(int ind, int sum) {
  if (sum < 0) return 0;
  if (ind == st.size()) {
    if (sum == 0)
      return 1;
    else
      return 0;
  }
  if (memo[ind][sum] != -1) return memo[ind][sum];
  int node = st[ind];
  int c1 = dp(ind + 1, sum - cnt[node][0]);
  int c2 = dp(ind + 1, sum - cnt[node][1]);
  if (c1 != 0) dir[ind][sum] = 0;
  if (c2 != 0) dir[ind][sum] = 1;
  return memo[ind][sum] = (c1 || c2);
}
int main() {
  cin >> n >> m;
  cin >> a >> b >> c;
  memset(cnt, 0, sizeof(cnt));
  memset(memo, -1, sizeof(memo));
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  x = a + c;
  y = b;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      st.push_back(i);
      dfs(i, 0, i);
    }
  }
  if (no or !dp(0, b)) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    int sum = b;
    for (int i = 0; i < st.size(); i++) {
      for (int j : con[st[i]]) {
        if (color[j] == dir[i][sum] and b > 0) {
          color[j] = 2;
          --b;
        } else {
          if (a > 0) {
            color[j] = 1;
            --a;
          } else {
            color[j] = 3;
            --c;
          }
        }
      }
      sum -= cnt[st[i]][dir[i][sum]];
    }
    if (a != 0) cout << a << endl;
    for (int i = 1; i <= n; i++) cout << color[i];
    cout << endl;
  }
}

```