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 MAXN = 200002;
int n, m, k;
int assigned_rank[MAXN];
vector<int> adj[MAXN];
int min_rank[MAXN];
int topo_order[MAXN];
int active_orders[MAXN];
set<int> people_by_rank[MAXN];
priority_queue<pair<int, int>> pq;
int countv;
int visited[MAXN];
bool topodfs(int v) {
  if (visited[v] == 2) return true;
  if (visited[v] == 1) return false;
  visited[v] = 1;
  for (int u : adj[v])
    if (!topodfs(u)) return false;
  topo_order[++countv] = v;
  visited[v] = 2;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> assigned_rank[i];
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    active_orders[y]++;
  }
  countv = 0;
  for (int i = 1; i <= n; i++)
    if (0 == visited[i])
      if (!topodfs(i)) {
        cout << "-1\n";
        return 0;
      }
  for (int i = 1; i <= n; i++) {
    int v = topo_order[i];
    min_rank[v] = 1;
    for (int u : adj[v]) min_rank[v] = max(min_rank[v], 1 + min_rank[u]);
    if (assigned_rank[v] > 0) {
      if (min_rank[v] > assigned_rank[v]) {
        cout << "-1\n";
        return 0;
      }
      min_rank[v] = assigned_rank[v];
    }
    if (min_rank[v] > k) {
      cout << "-1\n";
      return 0;
    }
    people_by_rank[min_rank[v]].insert(v);
    if (active_orders[v] == 0 && assigned_rank[v] == 0)
      pq.emplace(min_rank[v], v);
  }
  for (int r = k; r > 0; r--) {
    if (people_by_rank[r].empty()) {
      while (!pq.empty()) {
        auto p = pq.top();
        pq.pop();
        int u = p.second;
        if (min_rank[u] > r) continue;
        people_by_rank[min_rank[u]].erase(u);
        people_by_rank[r].insert(u);
        min_rank[u] = r;
        break;
      }
      if (people_by_rank[r].empty()) {
        cout << "-1\n";
        return 0;
      }
    }
    for (int v : people_by_rank[r]) {
      for (int u : adj[v]) {
        active_orders[u]--;
        if (active_orders[u] == 0 && assigned_rank[u] == 0)
          pq.emplace(min_rank[u], u);
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << min_rank[i] << ' ';
  cout << '\n';
  return 0;
}