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;
priority_queue<pair<int, int> > h;
vector<int> g[200010], w[200010], s[200010];
int q[200010], d[200010], ans[200010], a[200010], v[200010], mi[200010], n, m,
    k, f, r;
void work(int x) {
  for (int j = 0; j < w[x].size(); j++) {
    int y = w[x][j];
    --d[y];
    if (d[y] == 0 && !ans[y]) h.push(make_pair(mi[y], y));
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ans[i] = a[i];
    s[a[i]].push_back(i);
    v[a[i]] = 1;
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[y].push_back(x);
    w[x].push_back(y);
    d[x]++;
  }
  f = 1, r = 0;
  for (int i = 1; i <= n; i++)
    if (!d[i]) {
      q[++r] = i;
      mi[i] = 1;
    }
  while (f <= r) {
    int x = q[f++];
    if (a[x]) {
      if (a[x] < mi[x]) {
        puts("-1");
        return 0;
      } else
        mi[x] = a[x];
    }
    for (int i = 0; i < g[x].size(); i++) {
      int y = g[x][i];
      mi[y] = max(mi[y], mi[x] + 1);
      if (--d[y] == 0) q[++r] = y;
    }
  }
  for (int i = 1; i <= n; i++) d[i] = g[i].size();
  for (int i = 1; i <= n; i++)
    if (!d[i] && !ans[i]) {
      h.push(make_pair(mi[i], i));
    }
  for (int i = k; i >= 1; i--) {
    for (int j = 0; j < s[i].size(); j++) work(s[i][j]);
    while (!h.empty()) {
      int x = h.top().second;
      if (mi[x] > i) {
        puts("-1");
        return 0;
      }
      if (mi[x] == i) {
        ans[x] = i;
        v[i] = 1;
        work(x);
        h.pop();
      } else
        break;
    }
    if (!v[i]) {
      if (!h.empty()) {
        int x = h.top().second;
        ans[x] = i;
        work(x);
        h.pop();
      } else {
        puts("-1");
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < g[i].size(); j++)
      if (ans[i] >= ans[g[i][j]]) {
        puts("-1");
        return 0;
      }
  for (int i = 1; i <= n; i++) {
    printf("%d", ans[i]);
    if (i != n)
      printf(" ");
    else
      puts("");
  }
}