#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 9;
const int MAX = 1e2 + 9;
const double ep = 1e-4;
const double ep1 = 1e-6;
const int mod = 77797;
const double inf = 1e20;
const double pi = acos(-1);
int n, m, n1, n2, n3, cnt;
vector<int> vec[maxn], vec1[maxn];
int vis[maxn], sum[maxn][2], ans[maxn];
bool judge, dp[maxn][maxn][2];
void dfs(int cur, int pre, bool ju) {
  if (judge) return;
  vec1[cnt].emplace_back(cur);
  if (ju)
    sum[cnt][1]++, vis[cur] = 1;
  else
    vis[cur] = -1, sum[cnt][0]++;
  for (int j : vec[cur]) {
    if (vis[j]) {
      if (vis[j] == vis[cur]) {
        judge = 1;
        return;
      }
    } else
      dfs(j, cur, ju ^ 1);
  }
}
void solve() {
  scanf("%d%d", &(n), &(m));
  scanf("%d%d%d", &n1, &n2, &n3);
  for (int i = (1); i <= (m); ++i) {
    int u, v;
    scanf("%d%d", &(u), &(v));
    vec[u].emplace_back(v);
    vec[v].emplace_back(u);
  }
  for (int i = (1); i <= (n); ++i)
    if (!vis[i]) cnt++, dfs(i, -1, 1);
  if (judge) {
    puts("NO");
    return;
  }
  dp[0][0][0] = dp[0][0][1] = 1;
  for (int i = (1); i <= (cnt); ++i) {
    for (int j = (0); j <= (n); ++j) {
      if (dp[i - 1][j][0] || dp[i - 1][j][1]) {
        dp[i][j + sum[i][1]][1] = dp[i][j + sum[i][0]][0] = 1;
      }
    }
  }
  if (!dp[cnt][n2][0] && !dp[cnt][n2][1]) {
    puts("NO");
    return;
  }
  int cur = n2;
  puts("YES");
  for (int i = (cnt); i >= (1); --i) {
    if (dp[i - 1][cur - sum[i][1]][0] || dp[i - 1][cur - sum[i][1]][1]) {
      cur -= sum[i][1];
      for (int j : vec1[i]) {
        if (vis[j] == 1)
          ans[j] = 2;
        else if (n1)
          ans[j] = 1, n1--;
        else
          ans[j] = 3;
      }
      continue;
    }
    cur -= sum[i][0];
    for (int j : vec1[i]) {
      if (vis[j] == -1)
        ans[j] = 2;
      else if (n1)
        ans[j] = 1, n1--;
      else
        ans[j] = 3;
    }
  }
  for (int i = (1); i <= (n); ++i) printf("%d", ans[i]);
  puts("");
}
int main() {
  int T = 1;
  while (T--) solve();
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int maxn = 5e3 + 9;
const int MAX = 1e2 + 9;
const double ep = 1e-4;
const double ep1 = 1e-6;
const int mod = 77797;
const double inf = 1e20;
const double pi = acos(-1);
int n, m, n1, n2, n3, cnt;
vector<int> vec[maxn], vec1[maxn];
int vis[maxn], sum[maxn][2], ans[maxn];
bool judge, dp[maxn][maxn][2];
void dfs(int cur, int pre, bool ju) {
  if (judge) return;
  vec1[cnt].emplace_back(cur);
  if (ju)
    sum[cnt][1]++, vis[cur] = 1;
  else
    vis[cur] = -1, sum[cnt][0]++;
  for (int j : vec[cur]) {
    if (vis[j]) {
      if (vis[j] == vis[cur]) {
        judge = 1;
        return;
      }
    } else
      dfs(j, cur, ju ^ 1);
  }
}
void solve() {
  scanf("%d%d", &(n), &(m));
  scanf("%d%d%d", &n1, &n2, &n3);
  for (int i = (1); i <= (m); ++i) {
    int u, v;
    scanf("%d%d", &(u), &(v));
    vec[u].emplace_back(v);
    vec[v].emplace_back(u);
  }
  for (int i = (1); i <= (n); ++i)
    if (!vis[i]) cnt++, dfs(i, -1, 1);
  if (judge) {
    puts("NO");
    return;
  }
  dp[0][0][0] = dp[0][0][1] = 1;
  for (int i = (1); i <= (cnt); ++i) {
    for (int j = (0); j <= (n); ++j) {
      if (dp[i - 1][j][0] || dp[i - 1][j][1]) {
        dp[i][j + sum[i][1]][1] = dp[i][j + sum[i][0]][0] = 1;
      }
    }
  }
  if (!dp[cnt][n2][0] && !dp[cnt][n2][1]) {
    puts("NO");
    return;
  }
  int cur = n2;
  puts("YES");
  for (int i = (cnt); i >= (1); --i) {
    if (dp[i - 1][cur - sum[i][1]][0] || dp[i - 1][cur - sum[i][1]][1]) {
      cur -= sum[i][1];
      for (int j : vec1[i]) {
        if (vis[j] == 1)
          ans[j] = 2;
        else if (n1)
          ans[j] = 1, n1--;
        else
          ans[j] = 3;
      }
      continue;
    }
    cur -= sum[i][0];
    for (int j : vec1[i]) {
      if (vis[j] == -1)
        ans[j] = 2;
      else if (n1)
        ans[j] = 1, n1--;
      else
        ans[j] = 3;
    }
  }
  for (int i = (1); i <= (n); ++i) printf("%d", ans[i]);
  puts("");
}
int main() {
  int T = 1;
  while (T--) solve();
}

```