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
#include <bits/stdc++.h>
using namespace std;
vector<int> v[5010], sg, cc_rt;
int n1, n2, n3;
int dead1[5010], sz[5010], dep[5010], hv_odd = 0;
int dead2[5010], cntA[5010], cntB[5010], ans[5010];
int dead3[5010], col_rt[5010];
bool dp[5010][5010], pv[5010][5010];
void dfs1(int now, int rt) {
  dead1[now] = 1;
  sz[rt]++;
  for (auto x : v[now]) {
    if (!dead1[x]) {
      dep[x] = dep[now] + 1;
      dfs1(x, rt);
    } else if ((dep[now] + dep[x] + 1) & 1)
      hv_odd = 1;
  }
}
void dfs2(int now, int rt, int col = 0) {
  dead2[now] = 1;
  if (col)
    cntB[rt]++;
  else
    cntA[rt]++;
  for (auto x : v[now])
    if (!dead2[x]) dfs2(x, rt, 1 - col);
}
void dfs3(int now, int col) {
  if (col == 0)
    ans[now] = 2;
  else {
    if (n1)
      ans[now] = 1, n1--;
    else
      ans[now] = 3, n3--;
  }
  dead3[now] = 1;
  for (auto x : v[now])
    if (!dead3[x]) dfs3(x, 1 - col);
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  cin >> n1 >> n2 >> n3;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (!dead1[i]) dfs1(i, i);
  if (hv_odd) {
    cout << "NO\n";
    return 0;
  }
  cc_rt.push_back(-1);
  for (int i = 1; i <= n; i++)
    if (sz[i] > 1) {
      dfs2(i, i);
      cc_rt.push_back(i);
    }
  for (int i = 1; i <= n; i++)
    if (sz[i] == 1) sg.push_back(i);
  dp[0][0] = 1;
  for (int i = 1; i < cc_rt.size(); i++) {
    int now = cc_rt[i];
    for (int j = cntA[now]; j <= n2; j++) {
      if (dp[i - 1][j - cntA[now]]) {
        dp[i][j] = 1;
        pv[i][j] = 0;
      }
    }
    for (int j = cntB[now]; j <= n2; j++) {
      if (dp[i - 1][j - cntB[now]]) {
        dp[i][j] = 1;
        pv[i][j] = 1;
      }
    }
  }
  int pp = cc_rt.size() - 1, sg_sz = sg.size();
  for (int i = 0; i <= n2; i++) {
    if (dp[pp][i] == 1 && i + sg_sz >= n2) {
      cout << "YES\n";
      int val = i;
      for (int j = pp; j >= 0; j--) {
        if (pv[j][val] == 0) {
          col_rt[cc_rt[j]] = 0;
          val -= cntA[cc_rt[j]];
        } else {
          col_rt[cc_rt[j]] = 1;
          val -= cntB[cc_rt[j]];
        }
      }
      for (int j = 1; j <= n; j++)
        if (sz[j] > 1) dfs3(j, col_rt[j]);
      int cnt_2 = n2 - i;
      for (int j = 0; j < sg_sz; j++) {
        if (cnt_2) {
          ans[sg[j]] = 2;
          cnt_2--;
        } else {
          if (n1) {
            ans[sg[j]] = 1;
            n1--;
          } else {
            ans[sg[j]] = 3;
            n3--;
          }
        }
      }
      for (int j = 1; j <= n; j++) cout << ans[j];
      cout << endl;
      return 0;
    }
  }
  cout << "NO\n";
}
