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, d[300000], dr[300000], r[300000], rr[300000], lr[300000];
vector<int> e[300000], er[300000];
vector<int> ver[300000];
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> r[i];
    if (r[i] == 0)
      lr[i] = 1, rr[i] = k;
    else
      lr[i] = rr[i] = r[i];
  }
  int x, y;
  for (int i = 0; i < m; i++)
    cin >> x >> y, x--, y--, e[x].push_back(y), d[y]++, er[y].push_back(x),
        dr[x]++;
  queue<int> q;
  for (int i = 0; i < n; i++)
    if (d[i] == 0) q.push(i);
  int c = 0;
  while (!q.empty()) {
    int i = q.front();
    c++, q.pop();
    for (int j : e[i]) {
      rr[j] = min(rr[j], rr[i] - 1), d[j]--;
      if (d[j] == 0) q.push(j);
    }
  }
  for (int i = 0; i < n; i++)
    if (dr[i] == 0) q.push(i);
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    for (int j : er[i]) {
      lr[j] = max(lr[j], lr[i] + 1), dr[j]--;
      if (dr[j] == 0) q.push(j);
    }
  }
  bool pos = c == n && k <= n;
  for (int i = 0; i < n; i++) pos = pos && lr[i] <= rr[i];
  if (pos)
    for (int i = 0; i < n; i++) ver[rr[i]].push_back(i);
  if (pos) {
    priority_queue<pair<int, int>> pq;
    c = k;
    for (int i : ver[c]) pq.push({lr[i], i});
    while (!pq.empty()) {
      pair<int, int> p = pq.top();
      pq.pop();
      int i = p.second;
      if (rr[i] < c) {
        pos = false;
        break;
      }
      r[i] = max(lr[i], c);
      if (r[i] == c) {
        c--;
        if (c > 0)
          for (int i : ver[c]) pq.push({lr[i], i});
      }
    }
  }
  pos = pos && c == 0;
  if (!pos)
    cout << -1 << "\n";
  else
    for (int i = 0; i < n; i++) cout << r[i] << "\n";
}