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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
std::vector<int> topologicalOrder(const std::vector<std::vector<int>> &graph) {
  int n = (int)graph.size();
  std::vector<int> deg(n, 0);
  for (int i = 0; i < n; i++) {
    for (auto j : graph[i]) {
      deg[j]++;
    }
  }
  std::vector<int> order;
  for (int i = 0; i < n; i++) {
    if (deg[i] == 0) {
      order.push_back(i);
    }
  }
  for (int i = 0; i < (int)order.size(); i++) {
    int on = order[i];
    for (auto to : graph[on]) {
      if (--deg[to] == 0) {
        order.push_back(to);
      }
    }
  }
  return order;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n, m, k;
  std::cin >> n >> m >> k;
  std::vector<int> rank(n), mx(n, k), mn(n, 1);
  std::vector<std::vector<int>> graph(n);
  for (int i = 0; i < n; i++) {
    std::cin >> rank[i];
    if (rank[i]) {
      rank[i] = k - rank[i] + 1;
    }
  }
  while (m--) {
    int u, v;
    std::cin >> u >> v;
    u--;
    v--;
    graph[u].push_back(v);
  }
  auto ord = topologicalOrder(graph);
  if ((int)ord.size() != n) {
    std::cout << "-1\n";
    return 0;
  }
  for (auto i : ord) {
    if (rank[i]) {
      if (rank[i] < mn[i]) {
        std::cout << "-1\n";
        return 0;
      }
      mn[i] = rank[i];
    }
    if (mn[i] > k) {
      std::cout << "-1\n";
      return 0;
    }
    for (auto j : graph[i]) {
      mn[j] = std::max(mn[j], mn[i] + 1);
    }
  }
  std::reverse(ord.begin(), ord.end());
  for (auto i : ord) {
    for (auto j : graph[i]) {
      mx[i] = std::min(mx[i], mx[j] - 1);
    }
    if (rank[i]) {
      assert(mx[i] >= rank[i]);
      mx[i] = rank[i];
    }
    assert(mn[i] <= mx[i]);
  }
  std::reverse(ord.begin(), ord.end());
  std::vector<std::vector<int>> events(k + 1);
  std::vector<int> rev(n);
  for (int i = 0; i < n; i++) {
    rev[ord[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    events[mx[i]].push_back(i);
  }
  std::priority_queue<std::pair<int, int>> hp;
  for (int i = k; i > 0; i--) {
    for (auto id : events[i]) {
      hp.push({mn[id], rev[id]});
    }
    if (hp.empty()) {
      std::cout << "-1\n";
      return 0;
    }
    do {
      rank[ord[hp.top().second]] = i;
      hp.pop();
    } while (!hp.empty() && hp.top().first == i);
  }
  for (int i = 0; i < n; i++) {
    std::cout << k - rank[i] + 1 << (i + 1 == n ? '\n' : ' ');
  }
}