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;
vector<int> v[1000001];
int degree[1000001];
int ans[1000001];
bool toposort(int n) {
  int k = 0;
  for (int i = 1; i <= n; i++) {
    for (int j : v[i]) degree[j]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!degree[i]) ans[k++] = i;
  }
  for (int j = 0; j < k; j++) {
    for (int i : v[ans[j]]) {
      if (!--degree[i]) {
        ans[k++] = i;
      }
    }
  }
  return k == n;
}
int a[200001], l[200001], r[200001];
int order[200001];
int cnum;
bool cmp(int i, int j) { return l[i] > l[j]; }
bool solve(int n) {
  set<int> color;
  for (int t = n - 1; t >= 0; t--) {
    int i = ans[t];
    l[i] = 1;
    r[i] = cnum;
    for (int j : v[i]) l[i] = max(l[i], l[j] + 1);
    if (a[i] && a[i] < l[i]) return false;
    l[i] = max(l[i], a[i]);
  }
  for (int i = 1; i <= cnum; i++) color.insert(i);
  for (int i = 1; i <= n; i++) {
    color.erase(a[i]);
    order[i] = i;
  }
  sort(order + 1, order + n + 1, cmp);
  for (int k = 1; k <= n; k++) {
    int i = order[k];
    if (r[i] < l[i]) return false;
    if (!a[i]) {
      auto it = color.upper_bound(r[i]);
      if (it == color.begin() || *--it < l[i])
        a[i] = r[i];
      else {
        a[i] = *it;
        color.erase(it);
      }
    }
    for (int j : v[i]) r[j] = min(r[j], a[i] - 1);
  }
  return color.empty();
}
int main() {
  int n, m, x, y;
  scanf("%d%d%d", &n, &m, &cnum);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
  }
  if (!toposort(n) || !solve(n))
    printf("-1");
  else
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
}