Problem: You are given a connected undirected graph with n vertices and m edges. The vertices are enumerated from 1 to n. 

You are given n integers c1, c2, ..., cn, each of them is between  - n and n, inclusive. It is also guaranteed that the parity of cv equals the parity of degree of vertex v. The degree of a vertex is the number of edges connected to it.

You are to write a weight between  - 2·n2 and 2·n2 (inclusive) on each edge in such a way, that for each vertex v the sum of weights on edges connected to this vertex is equal to cv, or determine that this is impossible.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, n - 1 ≤ m ≤ 105) — the number of vertices and the number of edges.

The next line contains n integers c1, c2, ..., cn ( - n ≤ ci ≤ n), where ci is the required sum of weights of edges connected to vertex i. It is guaranteed that the parity of ci equals the parity of degree of vertex i.

The next m lines describe edges of the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), meaning that the i-th edge connects vertices ai and bi.

It is guaranteed that the given graph is connected and does not contain loops and multiple edges.

Output

If there is no solution, print "NO".

Otherwise print "YES" and then m lines, the i-th of them is the weight of the i-th edge wi ( - 2·n2 ≤ wi ≤ 2·n2).

Examples

Input

3 3
2 2 2
1 2
2 3
1 3


Output

YES
1
1
1


Input

4 3
-1 0 2 1
1 2
2 3
3 4


Output

YES
-1
1
1


Input

6 6
3 5 5 5 1 5
1 4
3 2
4 3
4 5
3 5
5 6


Output

YES
3
5
3
-1
-3
5


Input

4 4
4 4 2 4
1 2
2 3
3 4
4 1


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 100007;
struct eT {
  void setd(int _u, int _v, int _l) { u = _u, v = _v, last = _l; }
  int u, v, last;
} edge[MAXN * 2];
int n, m;
int ke, la[MAXN];
long long c[MAXN];
bool vis[MAXN];
long long ans[MAXN];
int si[MAXN];
bool ins[MAXN];
int stk[MAXN], top;
int kc, cyc[MAXN], ce[MAXN];
void init();
void input();
void work();
bool findcyc(int now, int fa);
long long dfs(int now, int fe);
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &c[i]);
  ke = 0;
  memset(la, -1, sizeof(la));
  int u, v;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &u, &v);
    edge[ke].setd(u, v, la[u]);
    la[u] = ke++;
    edge[ke].setd(v, u, la[v]);
    la[v] = ke++;
  }
}
void work() {
  if (findcyc(1, -1)) {
    memset(vis, false, sizeof(vis));
    dfs(cyc[1], -1);
    long long del = c[cyc[1]] / 2;
    cyc[kc + 1] = cyc[1];
    for (int i = 1; i <= kc; ++i) {
      int now = cyc[i];
      for (int j = la[now]; ~j; j = edge[j].last) {
        if (edge[j].v == cyc[i + 1]) {
          if (i & 1)
            ans[j >> 1] += del;
          else
            ans[j >> 1] -= del;
        }
      }
    }
  } else {
    memset(vis, false, sizeof(vis));
    dfs(1, -1);
    if (c[1]) {
      printf("NO\n");
      return;
    }
  }
  printf("YES\n");
  for (int i = 0; i < m; ++i) printf("%I64d\n", ans[i]);
}
bool findcyc(int now, int fa) {
  if (vis[now]) {
    if (ins[now] && ((top - si[now]) & 1)) {
      kc = top - si[now];
      for (int i = 1; i <= kc; ++i) cyc[i] = stk[si[now] + i - 1];
      return true;
    } else
      return false;
  }
  vis[now] = true;
  ins[now] = true;
  si[now] = top;
  stk[top++] = now;
  for (int i = la[now]; ~i; i = edge[i].last) {
    if (edge[i].v ^ fa) {
      if (findcyc(edge[i].v, now)) return true;
    }
  }
  ins[now] = false;
  --top;
  return false;
}
long long dfs(int now, int fe) {
  if (vis[now]) return 0;
  vis[now] = true;
  for (int i = la[now]; ~i; i = edge[i].last) {
    c[now] -= dfs(edge[i].v, i >> 1);
  }
  if (~fe) {
    long long ret = c[now];
    ans[fe] += c[now];
    c[now] = 0;
    return ret;
  } else
    return 0;
}