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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, K, rk[200000 + 5], in[200000 + 5], q[200000 + 5], top, mx[200000 + 5],
    mn[200000 + 5], b[200000 + 5];
vector<int> v[200000 + 5], V[200000 + 5];
vector<pair<int, int> > s[200000 + 5];
priority_queue<pair<int, int> > Q;
int main() {
  n = read();
  m = read();
  K = read();
  if (K > n) return 0 * puts("-1");
  for (int i = 1; i <= n; ++i)
    b[rk[i] = read()] = 1, mx[i] = rk[i] ? rk[i] : K, mn[i] = rk[i] ? rk[i] : 1;
  for (int i = 1, j, k; i <= m; ++i)
    j = read(), k = read(), v[j].push_back(k), V[k].push_back(j), ++in[k];
  for (int i = 1; i <= n; ++i)
    if (!in[i]) q[++top] = i;
  for (int i = 1; i <= top; ++i)
    for (int j = 0; j < v[q[i]].size(); ++j)
      if (!--in[v[q[i]][j]]) q[++top] = v[q[i]][j];
  if (top != n) return 0 * puts("-1");
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < v[q[i]].size(); ++j)
      mx[v[q[i]][j]] = min(mx[v[q[i]][j]], mx[q[i]] - 1);
  for (int i = n; i; --i)
    for (int j = 0; j < V[q[i]].size(); ++j)
      mn[V[q[i]][j]] = max(mn[V[q[i]][j]], mn[q[i]] + 1);
  for (int i = 1; i <= n; ++i)
    if (mx[i] < mn[i])
      return 0 * puts("-1");
    else
      s[mx[i]].push_back(make_pair(mn[i], i));
  for (int i = K; i; --i) {
    for (int j = 0; j < s[i].size(); ++j) Q.push(s[i][j]);
    if (Q.empty()) return 0 * puts("-1");
    int x = Q.top().second;
    Q.pop();
    rk[x] = i;
    while (!Q.empty() && Q.top().first == i) rk[Q.top().second] = i, Q.pop();
  }
  for (int i = 1; i <= n; ++i) printf("%d ", rk[i]);
  return 0;
}