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[200001], t[200001], V;
int d[200001], a[200001], n, m, k, q[200001], l, r, x, fa[200011];
bool bo[200001];
int ask(int x) { return (fa[x] == x ? x : fa[x] = ask(fa[x])); }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), bo[i] = (a[i] > 0);
  for (int i = 1; i <= m; i++) scanf("%d%d", &l, &r), t[l].push_back(r), d[r]++;
  l = 1, r = 0;
  for (int i = 1; i <= n; i++)
    if (!d[i]) q[++r] = i, a[i] = ((!bo[i]) ? k : a[i]);
  while (l <= r) {
    x = q[l];
    for (int j = 0, siz = t[x].size(); j < siz; j++) {
      if (!a[t[x][j]] || a[t[x][j]] >= a[x]) {
        if (bo[t[x][j]] || a[x] == 1) return puts("-1"), 0;
        a[t[x][j]] = a[x] - 1;
      }
      if (!(--d[t[x][j]])) q[++r] = t[x][j];
    }
    l++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i]) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) v[a[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    V.clear();
    fa[i] = i;
    for (int j = 0, siz = v[i].size(); j < siz; j++) {
      x = v[i][j], r = 0;
      if (bo[x]) {
        V.push_back(x);
        continue;
      }
      for (int k = 0, sz = t[x].size(); k < sz; k++) r = max(r, a[t[x][k]]);
      r = ask(r + 1);
      if (r < i)
        a[x] = r, v[r].push_back(x), fa[r] = r + 1;
      else
        V.push_back(x);
    }
    v[i] = V;
    fa[i] = (v[i].size() ? i + 1 : i);
  }
  for (int i = 1; i <= k; i++)
    if (!v[i].size()) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}