Problem: There are n military men in the Berland army. Some of them have given orders to other military men by now. Given m pairs (xi, yi), meaning that the military man xi gave the i-th order to another military man yi.

It is time for reform! The Berland Ministry of Defence plans to introduce ranks in the Berland army. Each military man should be assigned a rank — integer number between 1 and k, inclusive. Some of them have been already assigned a rank, but the rest of them should get a rank soon.

Help the ministry to assign ranks to the rest of the army so that:

  * for each of m orders it is true that the rank of a person giving the order (military man xi) is strictly greater than the rank of a person receiving the order (military man yi); 
  * for each rank from 1 to k there is at least one military man with this rank. 

Input

The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105, 1 ≤ k ≤ 2·105) — number of military men in the Berland army, number of orders and number of ranks.

The second line contains n integers r1, r2, ..., rn, where ri > 0 (in this case 1 ≤ ri ≤ k) means that the i-th military man has been already assigned the rank ri; ri = 0 means the i-th military man doesn't have a rank yet.

The following m lines contain orders one per line. Each order is described with a line containing two integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi). This line means that the i-th order was given by the military man xi to the military man yi. For each pair (x, y) of military men there could be several orders from x to y.

Output

Print n integers, where the i-th number is the rank of the i-th military man. If there are many solutions, print any of them.

If there is no solution, print the only number -1.

Examples

Input

5 3 3
0 3 0 0 2
2 4
3 4
3 5


Output

1 3 3 2 2 


Input

7 6 5
0 4 5 4 1 0 0
6 1
3 6
3 1
7 5
7 1
7 4


Output

2 4 5 4 1 3 5 


Input

2 2 2
2 1
1 2
2 1


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, M, K;
int R[200200];
vector<int> G[200200];
bool visited[200200];
int minR[200200];
int in_deg[200200];
bool ok = true;
int dfs(int n) {
  if (visited[n]) {
    return minR[n];
  }
  visited[n] = true;
  for (int x : G[n]) {
    minR[n] = max(minR[n], dfs(x));
    if (!ok) {
      return -1;
    }
  }
  ++minR[n];
  if (R[n] != 0 && minR[n] > R[n]) {
    ok = false;
    return -1;
  }
  return minR[n] = max(minR[n], R[n]);
}
int main() {
  cin >> N >> M >> K;
  for (int n = 0; n < N; ++n) {
    cin >> R[n];
  }
  while (M--) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].push_back(b);
    ++in_deg[b];
  }
  memset(visited, 0, sizeof(visited));
  vector<priority_queue<pair<int, int>>> Q(2);
  for (int n = 0; n < N && ok; ++n) {
    if (!visited[n]) {
      dfs(n);
    }
  }
  if (!ok) {
    cout << -1 << '\n';
    return 0;
  }
  for (int n = 0; n < N; ++n) {
    if (in_deg[n] == 0) {
      Q[R[n] == 0].emplace(minR[n], n);
    }
  }
  auto remove = [&](priority_queue<pair<int, int>>& q) {
    int n = q.top().second;
    q.pop();
    for (int x : G[n]) {
      if (--in_deg[x] == 0) {
        Q[R[x] == 0].emplace(minR[x], x);
      }
    }
  };
  while (K > 0) {
    int cnt = 0;
    for (auto& q : Q) {
      while (q.size() && q.top().first >= K) {
        int n = q.top().second;
        if (minR[n] > K) {
          cout << -1 << '\n';
          return 0;
        }
        R[n] = K;
        ++cnt;
        remove(q);
      }
    }
    if (cnt == 0) {
      if (Q[1].empty()) {
        cout << -1 << '\n';
        return 0;
      }
      int n = Q[1].top().second;
      R[n] = K;
      remove(Q[1]);
    }
    --K;
  }
  if (count(R, R + N, 0)) {
    cout << -1 << '\n';
    return 0;
  }
  for (int n = 0; n < N; ++n) {
    cout << R[n] << ' ';
  }
  cout << '\n';
}