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>
const int N = 200005;
int n, m, k, ans[N], deg[N];
int min[N], max[N], u[N], v[N];
std::vector<int> g[N];
void link(int x, int y) { g[x].emplace_back(y), ++deg[y]; }
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n >> m >> k;
  for (int i = 1, x; i <= n; ++i) {
    std::cin >> x, min[i] = 1, max[i] = k;
    if (x) min[i] = max[i] = x;
  }
  for (int i = 1; i <= m; ++i) std::cin >> u[i] >> v[i], link(u[i], v[i]);
  std::queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) q.push(i);
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int y : g[x]) {
      max[y] = std::min(max[y], max[x] - 1);
      if (!--deg[y]) q.push(y);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (deg[i] || min[i] > max[i]) return std::cout << "-1\n", 0;
  for (int i = 1; i <= n; ++i) g[i].clear();
  for (int i = 1; i <= m; ++i) link(v[i], u[i]);
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) q.push(i);
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int y : g[x]) {
      min[y] = std::max(min[y], min[x] + 1);
      if (!--deg[y]) q.push(y);
    }
  }
  for (int i = 1; i <= n; ++i) g[i].clear();
  for (int i = 1; i <= n; ++i) g[min[i]].emplace_back(i);
  std::priority_queue<std::pair<int, int>> pq;
  for (int i = 1; i <= k; ++i) {
    for (int j : g[i]) pq.emplace(-max[j], j);
    if (pq.empty()) return std::cout << "-1\n", 0;
    do {
      int x = pq.top().second;
      pq.pop(), ans[x] = i;
    } while (pq.size() && pq.top().first == -i);
  }
  for (int i = 1; i <= n; ++i) std::cout << ans[i] << ' ';
  std::cout << '\n';
  return 0;
}