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, len, ll, rr, lr[1000000], nex[1000000], wen[1000000], hea[1000000],
    x[1000000], y[1000000], a[1000000], l[1000000], r[1000000], dig[1000000],
    ans[1000000];
vector<int> v[1000000];
priority_queue<pair<int, int>> p;
void add(int x, int y) {
  ++len;
  nex[len] = hea[x];
  wen[len] = y;
  hea[x] = len;
  ++dig[y];
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!a[i])
      l[i] = 1, r[i] = k;
    else
      l[i] = r[i] = a[i];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x[i], &y[i]);
    add(x[i], y[i]);
  }
  for (int i = 1; i <= n; i++)
    if (!dig[i]) lr[++rr] = i;
  while (ll < rr) {
    ++ll;
    for (int i = hea[lr[ll]]; i; i = nex[i]) {
      --dig[wen[i]];
      r[wen[i]] = min(r[wen[i]], r[lr[ll]] - 1);
      if (!dig[wen[i]]) lr[++rr] = wen[i];
    }
  }
  for (int i = 1; i <= n; i++)
    if (dig[i]) {
      puts("-1");
      return 0;
    }
  len = ll = rr = 0;
  for (int i = 1; i <= n; i++) hea[i] = 0;
  for (int i = 1; i <= m; i++) add(y[i], x[i]);
  for (int i = 1; i <= n; i++)
    if (!dig[i]) lr[++rr] = i;
  while (ll < rr) {
    ++ll;
    for (int i = hea[lr[ll]]; i; i = nex[i]) {
      --dig[wen[i]];
      l[wen[i]] = max(l[wen[i]], l[lr[ll]] + 1);
      if (!dig[wen[i]]) lr[++rr] = wen[i];
    }
  }
  for (int i = 1; i <= n; i++)
    if (dig[i] > 0 || l[i] > r[i]) {
      puts("        -1");
      return 0;
    }
  for (int i = 1; i <= n; i++) v[r[i]].push_back(i);
  for (int i = k; i >= 1; i--) {
    for (int j = 0; j < v[i].size(); j++)
      p.push(make_pair(l[v[i][j]], v[i][j]));
    if (!p.size()) {
      puts("-1");
      return 0;
    }
    int x = (p.top()).second;
    ans[x] = i;
    p.pop();
    while (!p.empty()) {
      x = (p.top()).second;
      if (l[x] < i) break;
      ans[x] = i;
      p.pop();
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}