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 a[200010], ans[200010];
int n, m, k;
int vis[200010], minn[200010], maxx[200010], dx[200010], dn[200010];
vector<int> g0[200010], g1[200010], t[200010];
int read() {
  int ans = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  return ans;
}
int gx(int u) {
  if (vis[u]) {
    printf("-1\n");
    exit(0);
  }
  if (dx[u]) return maxx[u];
  vis[u] = 1;
  for (int i = 0; i < g1[u].size(); i++)
    maxx[u] = min(maxx[u], gx(g1[u][i]) - 1);
  vis[u] = 0;
  dx[u] = 1;
  return maxx[u];
}
int gn(int u) {
  if (vis[u]) {
    printf("-1\n");
    exit(0);
  }
  if (dn[u]) return minn[u];
  vis[u] = 1;
  for (int i = 0; i < g0[u].size(); i++)
    minn[u] = max(minn[u], gn(g0[u][i]) + 1);
  vis[u] = 0;
  dn[u] = 1;
  return minn[u];
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i])
      minn[i] = maxx[i] = a[i];
    else
      maxx[i] = k, minn[i] = 1;
  }
  for (int i = 1; i <= m; i++) {
    int x = read(), y = read();
    g0[x].push_back(y);
    g1[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (gx(i) < gn(i) || gx(i) < 1 || gn(i) > k) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= n; i++) t[gx(i)].push_back(i);
  priority_queue<pair<int, int> > q;
  while (q.size()) q.pop();
  for (int i = k; i; i--) {
    for (int j = 0; j < t[i].size(); j++)
      q.push(make_pair(gn(t[i][j]), t[i][j]));
    if (!q.size()) {
      printf("-1\n");
      return 0;
    }
    ans[q.top().second] = i;
    q.pop();
    while (q.size()) {
      int u = q.top().second;
      if (gn(u) < i) break;
      q.pop();
      ans[u] = i;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
}