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;
const int N = 200100;
int fa[N], n, m, K, a[N], val[N], d[N], q[N];
vector<int> g[N], vec[N], V;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void gofail() {
  cout << -1 << '\n';
  exit(0);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> K;
  for (int i = 1; i <= n; ++i) cin >> a[i], val[i] = a[i];
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    d[y]++;
  }
  int t = 0, w = 0;
  for (int i = 1; i <= n; ++i)
    if (!d[i]) {
      if (!val[i]) val[i] = K;
      q[++w] = i;
    }
  for (; t < w;) {
    int x = q[++t];
    for (int i : g[x]) {
      if (a[i] >= val[x] || val[x] == 1) gofail();
      if (!val[i] || val[i] >= val[x]) val[i] = val[x] - 1;
      if (!--d[i]) q[++w] = i;
    }
  }
  if (w != n) gofail();
  for (int i = 1; i <= n; ++i) vec[val[i]].push_back(i);
  for (int i = 1; i <= K; ++i) {
    V.clear();
    fa[i] = i;
    for (int j : vec[i]) {
      if (a[j]) {
        V.push_back(j);
        continue;
      }
      int mx = 0;
      for (int k : g[j]) mx = max(mx, val[k]);
      mx = find(mx + 1);
      if (mx < i)
        val[j] = mx, vec[mx].push_back(j), fa[mx] = mx + 1;
      else
        V.push_back(j);
    }
    swap(vec[i], V);
    fa[i] = i + !vec[i].empty();
  }
  for (int i = 1; i <= K; ++i)
    if (vec[i].empty()) gofail();
  for (int i = 1; i <= n; ++i) cout << val[i] << " \n"[i == n];
  return 0;
}