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 = 200005;
vector<int> order, g[N], g2[N];
int dg[N], dp[N], n, m, k, e[N];
int32_t main() {
  scanf("%d%d%d", &n, &m, &k);
  set<int> miss;
  for (int i = 1; i <= k; i++) miss.insert(i);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &e[i]);
    miss.erase(e[i]);
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[y].push_back(x);
    g2[x].push_back(y);
    dg[x]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!dg[i]) order.push_back(i);
  }
  for (int i = 0; i < (int)(order).size(); i++) {
    auto v = order[i];
    for (auto w : g[v]) {
      if (--dg[w] == 0) order.push_back(w);
    }
  }
  if ((int)(order).size() != n) {
    puts("-1");
    return 0;
  }
  reverse(begin(order), end(order));
  vector<pair<int, int> > ord;
  for (int i = 0; i < (int)(order).size(); i++) {
    int has = k;
    auto v = order[i];
    for (auto w : g[v]) {
      has = min(has, dp[w] - 1);
    }
    dp[v] = has;
    if (e[v]) {
      if (dp[v] < e[v]) {
        puts("-1");
        return 0;
      }
      dp[v] = e[v];
    }
    if (dp[v] < 1) {
      puts("-1");
      return 0;
    }
    if (!e[v]) {
      ord.push_back({dp[v], v});
    }
  }
  cout << endl;
  sort(begin(ord), end(ord));
  for (auto w : ord) {
    int la = 1;
    for (auto wk : g2[w.second]) {
      la = max(la, e[wk] + 1);
    }
    auto fa = miss.lower_bound(la);
    if (fa == end(miss)) {
      e[w.second] = w.first;
    } else {
      e[w.second] = min(w.first, *fa);
    }
    miss.erase(e[w.second]);
  }
  for (int i = 1; i <= n; i++) {
    for (auto w : g2[i]) {
      if (e[i] <= e[w]) {
        puts("-1");
        return 0;
      }
    }
  }
  if ((int)(miss).size()) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", e[i]);
  }
  printf("\n");
}