Problem: In the country there are n cities and m bidirectional roads between them. Each city has an army. Army of the i-th city consists of ai soldiers. Now soldiers roam. After roaming each soldier has to either stay in his city or to go to the one of neighboring cities by at moving along at most one road.

Check if is it possible that after roaming there will be exactly bi soldiers in the i-th city.

Input

First line of input consists of two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 200).

Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 100).

Next line contains n integers b1, b2, ..., bn (0 ≤ bi ≤ 100).

Then m lines follow, each of them consists of two integers p and q (1 ≤ p, q ≤ n, p ≠ q) denoting that there is an undirected road between cities p and q. 

It is guaranteed that there is at most one road between each pair of cities.

Output

If the conditions can not be met output single word "NO".

Otherwise output word "YES" and then n lines, each of them consisting of n integers. Number in the i-th line in the j-th column should denote how many soldiers should road from city i to city j (if i ≠ j) or how many soldiers should stay in city i (if i = j).

If there are several possible answers you may output any of them.

Examples

Input

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


Output

YES
1 0 0 0 
2 0 0 0 
0 5 1 0 
0 0 2 1 


Input

2 0
1 2
2 1


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename COST = double>
struct graph {
  vector<int> dest;
  vector<vector<int> > adj;
  int inv(int a) { return a ^ 0x1; }
  graph(int n = 0) {
    adj.resize(n);
    imb.resize(n);
  }
  int arc(int i, int j, long long u = 1, COST c = 0) {
    dest.push_back(j);
    adj[i].push_back(((int)(dest).size()) - 1);
    dest.push_back(i);
    adj[j].push_back(((int)(dest).size()) - 1);
    cap.push_back(u);
    cap.push_back(0);
    cost.push_back(c);
    cost.push_back(-c);
    return ((int)(dest).size()) - 2;
  }
  vector<long long> cap, flow;
  int orig(int a) { return dest[inv(a)]; }
  long long capres(int a) { return cap[a] - flow[a]; }
  vector<int> curAdj, d;
  bool MFbfs(int s, int t) {
    d.assign(((int)(adj).size()), INT_MAX / 2);
    curAdj = vector<int>(((int)(adj).size()));
    d[s] = 0;
    queue<int> Q;
    Q.push(s);
    while (!Q.empty()) {
      int u = Q.front();
      Q.pop();
      for (auto i : adj[u]) {
        int v = dest[i];
        if (capres(i) > 0 && d[v] == INT_MAX / 2) {
          d[v] = d[u] + 1;
          Q.push(v);
        }
      }
    }
    return d[t] != INT_MAX / 2;
  }
  long long MFdfs(int u, int t, long long f) {
    if (u == t) return f;
    for (int &i = curAdj[u]; i < adj[u].size(); ++i) {
      int ar = adj[u][i], v = dest[ar];
      if (d[v] != d[u] + 1 || capres(ar) == 0) continue;
      long long tmpF = MFdfs(v, t, min(f, capres(ar)));
      if (tmpF) {
        flow[ar] += tmpF;
        flow[inv(ar)] -= tmpF;
        return tmpF;
      }
    }
    return 0;
  }
  long long maxflow(int ini, int end) {
    flow.assign(((int)(dest).size()), 0);
    while (MFbfs(ini, end))
      while (MFdfs(ini, end, LLONG_MAX / 2))
        ;
    long long F = 0;
    for (int a : adj[ini]) F += flow[a];
    return F;
  }
  vector<int> mincut(int ini) {
    vector<int> ans;
    vector<bool> mark(((int)(adj).size()), false);
    mark[ini] = true;
    ans.push_back(ini);
    for (int pos = 0; pos < ans.size(); ++pos) {
      int x = ans[pos];
      for (int ar : adj[x])
        if (capres(ar) > 0) {
          int y = dest[ar];
          if (mark[y]) continue;
          ans.push_back(y);
          mark[y] = true;
        }
    }
    return ans;
  }
  vector<long long> imb;
  vector<COST> cost, pot;
  long long delta;
  COST rescost(int a) { return cost[a] + pot[orig(a)] - pot[dest[a]]; }
  bool MCFdijkstra() {
    priority_queue<pair<COST, pair<int, int> > > q;
    vector<int> ent(((int)(adj).size()), -2);
    vector<COST> dist(((int)(adj).size()), INFINITY);
    for (int u = 0; u < ((int)(adj).size()); ++u)
      if (imb[u] >= delta) q.push(make_pair(0, make_pair(u, -1)));
    while (!q.empty()) {
      int u = q.top().second.first, f = q.top().second.second;
      COST d = -q.top().first;
      q.pop();
      if (ent[u] != -2) continue;
      dist[u] = d;
      ent[u] = f;
      for (int a : adj[u])
        if (capres(a) >= delta)
          q.push(make_pair(-(dist[u] + rescost(a)), make_pair(dest[a], a)));
    }
    for (int u = 0; u < ((int)(adj).size()); ++u)
      if (ent[u] != -2 && imb[u] <= -delta) {
        for (int v = 0; v < ((int)(adj).size()); ++v) pot[v] += dist[v];
        for (int a = ent[u]; a != -1; a = ent[orig(a)]) {
          flow[a] += delta;
          flow[inv(a)] -= delta;
          imb[dest[a]] += delta;
          imb[orig(a)] -= delta;
        }
        return true;
      }
    return false;
  }
  COST mincostflow() {
    pot.resize(((int)(adj).size()));
    flow.resize(((int)(dest).size()));
    for (delta = 0x40000000; delta > 0; delta /= 2) {
      for (int a = 0; a < ((int)(dest).size()); ++a) {
        int u = orig(a), v = dest[a];
        if (capres(a) >= delta && cmp(rescost(a)) < 0) {
          imb[u] -= capres(a);
          imb[v] += capres(a);
          flow[inv(a)] -= capres(a);
          flow[a] += capres(a);
        }
      }
      while (MCFdijkstra())
        ;
    }
    COST C = 0;
    for (int a = 0; a < ((int)(dest).size()); ++a)
      if (flow[a] > 0) C += flow[a] * cost[a];
    return C;
  }
  vector<int> depth;
  vector<bool> artp, bridge;
  vector<int> least;
  int dfs_artpbridge(int u, int ent) {
    int nf = 0;
    for (int a : adj[u]) {
      int v = dest[a];
      if (depth[v] == -1) {
        least[v] = depth[v] = depth[u] + 1;
        dfs_artpbridge(v, a);
        nf++;
        if (least[v] >= depth[u]) {
          artp[u] = true;
          if (least[v] == depth[v]) bridge[a] = bridge[inv(a)] = true;
        } else
          least[u] = min(least[u], least[v]);
      } else if (inv(a) != ent)
        least[u] = min(least[u], depth[v]);
    }
    return nf;
  }
  void artpbridge() {
    artp = vector<bool>(((int)(adj).size()), false);
    bridge = vector<bool>(((int)(dest).size()), false);
    depth = vector<int>(((int)(adj).size()), -1);
    least = vector<int>(((int)(adj).size()), -1);
    for (int i = 0; i < ((int)(adj).size()); ++i)
      if (depth[i] == -1) {
        least[i] = depth[i] = 0;
        if (dfs_artpbridge(i, -1) < 2) artp[i] = false;
      }
  }
  vector<int> ord, rep;
  int transp(int a) { return (a & 0x1); }
  void dfs_topsort(int u) {
    for (int a : adj[u]) {
      int v = dest[a];
      if (!transp(a) && depth[v] == -1) {
        depth[v] = depth[u] + 1;
        dfs_topsort(v);
      }
    }
    ord.push_back(u);
  }
  void dfs_compfortcon(int u, int ent) {
    rep[u] = ent;
    for (int a : adj[u]) {
      int v = dest[a];
      if (transp(a) && rep[v] == -1) dfs_compfortcon(v, ent);
    }
  }
  void compfortcon() {
    depth = vector<int>(((int)(adj).size()), -1);
    ord.clear();
    for (int u = 0; u < ((int)(adj).size()); ++u)
      if (depth[u] == -1) {
        depth[u] = 0;
        dfs_topsort(u);
      }
    rep = vector<int>(((int)(adj).size()), -1);
    for (int i = ((int)(adj).size()) - 1; i >= 0; i--)
      if (rep[ord[i]] == -1) dfs_compfortcon(ord[i], ord[i]);
  }
  vector<int> match;
  int dfs_match(int node, int m) {
    if (match[node] == m) return 0;
    match[node] = m;
    for (int ar : adj[node]) {
      int j = dest[ar];
      if (match[j] == -1 || dfs_match(match[j], m)) {
        match[j] = node;
        return 1;
      }
    }
    return 0;
  }
  int bipartite_match(int k) {
    int N = ((int)(adj).size());
    match = vector<int>(N, -1);
    int ans = 0;
    for (int i = 0; i < k; ++i) ans += dfs_match(i, i);
    for (int i = 0; i < k; ++i) match[i] = -1;
    for (int i = k; i < N; ++i)
      if (match[i] != -1) match[match[i]] = i;
    return ans;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  int N, M;
  cin >> N >> M;
  graph<> g(2 * N + 2);
  long long a, b, totA = 0, totB = 0;
  for (int i = 1; i <= N; i++) {
    cin >> a;
    totA += a;
    g.arc(0, i, a);
  }
  for (int i = 1; i <= N; i++) {
    cin >> b;
    totB += b;
    g.arc(N + i, 2 * N + 1, b);
  }
  if (totA != totB) {
    cout << "NO\n";
    return 0;
  }
  long long tot = totA;
  for (int i = 1; i <= N; i++) {
    g.arc(i, N + i, LLONG_MAX / 2);
  }
  for (int e = 0; e < M; e++) {
    cin >> a >> b;
    g.arc(a, N + b, LLONG_MAX / 2);
    g.arc(b, N + a, LLONG_MAX / 2);
  }
  long long res = g.maxflow(0, 2 * N + 1);
  res = 0;
  vector<int> cut = g.mincut(0);
  vector<bool> isCut(g.adj.size(), false);
  for (int x : cut) isCut[x] = true;
  for (int e = 0; e < g.dest.size(); e++) {
    int u = g.orig(e), v = g.dest[e];
    if (isCut[u] && !isCut[v]) res += g.cap[e];
  }
  if (res != tot) {
    cout << "NO\n";
    return 0;
  }
  vector<vector<long long> > out(N, vector<long long>(N));
  for (int e = 0; e < g.dest.size(); ++e) {
    int u = g.orig(e), v = g.dest[e];
    long long f = g.flow[e];
    if (f > 0 && (1 <= u && u <= N) && (N + 1 <= v && v <= 2 * N)) {
      out[u - 1][v - N - 1] += f;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      cout << out[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}