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;
vector<int> adj[200010];
vector<int> adj2[200010];
vector<int> rev[200010];
vector<int> revprelvl[200010];
int deg[200010];
int deg2[200010];
int lvl[200010], prelvl[200010];
struct Jihoon {
  int x;
  Jihoon(int _x) { x = _x; }
  inline bool operator<(const Jihoon &c) const { return lvl[x] < lvl[c.x]; }
};
int main() {
  int v, e, k;
  scanf("%d%d%d", &v, &e, &k);
  for (int i = 1; i <= v; i++) {
    scanf("%d", &prelvl[i]);
    lvl[i] = prelvl[i];
    if (prelvl[i]) revprelvl[prelvl[i]].push_back(i);
  }
  int x, y;
  for (int i = 0; i < e; i++) {
    scanf("%d%d", &x, &y);
    adj[y].push_back(x);
    adj2[x].push_back(y);
    deg[x]++;
    deg2[y]++;
  }
  queue<int> q;
  for (x = 1; x <= v; x++) {
    if (deg[x] == 0) {
      q.push(x);
      lvl[x] = max(1, lvl[x]);
    }
  }
  int maxlvl = 0;
  for (int i = 0; i < v; i++) {
    if (q.empty()) {
      printf("-1");
      return 0;
    }
    x = q.front();
    q.pop();
    rev[lvl[x]].push_back(x);
    maxlvl = max(lvl[x], maxlvl);
    for (auto y : adj[x]) {
      deg[y]--;
      if (deg[y] == 0) q.push(y);
      lvl[y] = max(lvl[x] + 1, lvl[y]);
    }
  }
  if (maxlvl > k) {
    printf("-1");
    return 0;
  }
  for (x = 1; x <= v; x++) {
    if (prelvl[x]) {
      if (prelvl[x] < lvl[x]) {
        printf("-1");
        return 0;
      }
      lvl[x] = prelvl[x];
    }
  }
  priority_queue<struct Jihoon> pq;
  for (x = 1; x <= v; x++) {
    if (deg2[x] == 0 && !prelvl[x]) {
      pq.push(Jihoon(x));
    }
  }
  for (int ll = k; ll > 0; ll--) {
    bool flag = false;
    for (auto x : revprelvl[ll]) {
      for (auto y : adj2[x]) {
        deg2[y]--;
        if (!deg2[y] && !prelvl[y]) {
          pq.push(Jihoon(y));
        }
      }
      flag = true;
    }
    while (!flag || (!pq.empty() && lvl[pq.top().x] == ll)) {
      if (pq.empty() || lvl[pq.top().x] > ll) {
        printf("-1");
        return 0;
      }
      x = pq.top().x;
      pq.pop();
      for (auto y : adj2[x]) {
        deg2[y]--;
        if (!deg2[y] && !prelvl[y]) {
          pq.push(Jihoon(y));
        }
      }
      lvl[x] = ll;
      if (pq.empty() || lvl[pq.top().x] != ll) break;
    }
  }
  for (int i = 1; i <= v; i++) printf("%d ", lvl[i]);
  return 0;
}