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 = 2e5 + 10;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int n, m, k, deg[N], seq[N], tt, l[N], r[N], ans[N];
vector<int> E[N], vec[N];
void No() {
  cout << -1;
  exit(0);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    r[i] = gi();
    if (r[i])
      l[i] = r[i];
    else
      l[i] = 1, r[i] = k;
  }
  for (int i = 1, u, v; i <= m; i++)
    u = gi(), v = gi(), E[u].push_back(v), ++deg[v];
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    seq[++tt] = u;
    for (auto v : E[u])
      if (!--deg[v]) q.push(v);
  }
  if (tt != n) No();
  for (int i = 1; i <= n; i++) {
    int u = seq[i];
    for (auto v : E[u]) r[v] = min(r[v], r[u] - 1);
  }
  for (int i = n; i; i--) {
    int u = seq[i];
    for (auto v : E[u]) l[u] = max(l[u], l[v] + 1);
  }
  for (int i = 1; i <= n; i++) {
    if (l[i] > r[i]) No();
    vec[l[i]].push_back(i);
  }
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      Q;
  for (int i = 1; i <= k; i++) {
    for (auto t : vec[i]) Q.push(make_pair(r[t], t));
    if (Q.empty()) No();
    do {
      ans[Q.top().second] = i;
      Q.pop();
    } while (!Q.empty() && Q.top().first == i);
  }
  for (int u = 1; u <= n; u++)
    for (auto v : E[u]) assert(ans[u] > ans[v]);
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}