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, a[200010], q[200010], fa[200010], deg[200010];
bool b[200010];
vector<int> V, g[200010], vec[200010];
int get(int u) { return fa[u] == u ? u : fa[u] = get(fa[u]); }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = (a[i] > 0);
  for (int i = 1, u, v; i <= m; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    deg[v]++;
  }
  int l = 1, r = 0;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q[++r] = i, a[i] = ((!b[i]) ? k : a[i]);
  while (l <= r) {
    int u = q[l];
    for (auto v : g[u]) {
      if (!a[v] || a[v] >= a[u]) {
        if (b[v] || a[u] == 1) return puts("-1"), 0;
        a[v] = a[u] - 1;
      }
      if (!(--deg[v])) q[++r] = v;
    }
    l++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i]) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) vec[a[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    V.clear();
    fa[i] = i;
    for (auto u : vec[i]) {
      r = 0;
      if (b[u]) {
        V.push_back(u);
        continue;
      }
      for (auto v : g[u]) r = max(r, a[v]);
      r = get(r + 1);
      if (r < i)
        a[u] = r, vec[r].push_back(u), fa[r] = r + 1;
      else
        V.push_back(u);
    }
    vec[i] = V;
    fa[i] = (vec[i].size() ? i + 1 : i);
  }
  for (int i = 1; i <= k; i++)
    if (!vec[i].size()) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}