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;
const int N = 200005;
int n, m, k, a[N], b[N];
int deg[N], q[N], vis[N], fa[N];
vector<int> e[N], v[N];
int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(1); i <= (int)(n); i++) b[i] = (a[i] ? a[i] : k);
  for (int i = (int)(1); i <= (int)(m); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    ++deg[y];
  }
  int h = 0, t = 0;
  for (int i = (int)(1); i <= (int)(n); i++)
    if (!deg[i]) q[++t] = i;
  while (h != t) {
    int x = q[++h];
    if ((a[x] && b[x] < a[x]) || b[x] < 1) return puts("-1"), 0;
    for (auto i : e[x]) {
      b[i] = min(b[i], b[x] - 1);
      if (!--deg[i]) q[++t] = i;
    }
  }
  if (t != n) return puts("-1"), 0;
  for (int i = (int)(1); i <= (int)(n); i++) v[b[i]].push_back(i);
  for (int i = (int)(1); i <= (int)(k); i++) fa[i] = i;
  for (int i = (int)(1); i <= (int)(k); i++) {
    for (auto j : v[i]) {
      if (a[j]) {
        vis[i] = 1;
        continue;
      }
      int r = 0;
      for (auto k : e[j]) r = max(r, b[k]);
      r = get(r + 1);
      if (r != i) fa[r] = r + 1;
      b[j] = r;
      vis[r] = 1;
    }
    if (vis[i]) fa[i] = i + 1;
  }
  for (int i = (int)(1); i <= (int)(k); i++)
    if (!vis[i]) return puts("-1"), 0;
  for (int i = (int)(1); i <= (int)(n); i++) printf("%d ", b[i]);
}