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 std::cin;
using std::cout;
const int maxn = 200100;
const int inf = 1e9;
std::vector<int> v0[maxn], v1[maxn], nu[maxn];
int n, m, k;
int max[maxn], min[maxn], vis[maxn];
int ans[maxn], mx[maxn], mn[maxn];
inline int gmax(int x) {
  if (vis[x]) puts("-1"), exit(0);
  if (mx[x]) return max[x];
  vis[x] = 1;
  for (int i : v1[x]) max[x] = std::min(max[x], gmax(i) - 1);
  vis[x] = 0, mx[x] = 1;
  return max[x];
}
inline int gmin(int x) {
  if (vis[x]) puts("-1"), exit(0);
  if (mn[x]) return min[x];
  vis[x] = 1;
  for (int i : v0[x]) min[x] = std::max(min[x], gmin(i) + 1);
  vis[x] = 0, mn[x] = 1;
  return min[x];
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> max[i];
    if (!max[i]) {
      max[i] = k;
      min[i] = 1;
    } else {
      min[i] = max[i];
    }
  }
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    v0[x].push_back(y);
    v1[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (gmax(i) < gmin(i) || gmax(i) < 1 || gmin(i) > k) {
      puts("-1");
      return 0;
    }
  for (int i = 1; i <= n; ++i) nu[gmax(i)].push_back(i);
  std::priority_queue<std::pair<int, int>> q;
  for (int i = k; i >= 1; --i) {
    for (int x : nu[i]) q.emplace(gmin(x), x);
    if (q.empty()) {
      puts("-1");
      return 0;
    }
    ans[q.top().second] = i, q.pop();
    for (; q.size();) {
      int x = q.top().second;
      if (gmin(x) < i) break;
      q.pop(), ans[x] = i;
    }
  }
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << ' ';
  }
}