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 = 2e5 + 10;
int a[maxn];
vector<int> g[maxn];
set<int> v[maxn];
int dp[maxn], in[maxn];
bool del[maxn];
void impossible() {
  cout << -1 << endl;
  exit(0);
}
int dfs(int x) {
  int &ans = dp[x];
  if (ans != -1) return ans;
  ans = max(1, a[x]);
  for (auto i : g[x]) {
    ans = max(ans, 1 + dfs(i));
  }
  if (a[x] && ans != a[x]) impossible();
  return ans;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    int p, q;
    cin >> p >> q;
    g[p].push_back(q);
    in[q] += 1;
  }
  memset(dp, -1, sizeof dp);
  for (int i = 1; i <= n; i++)
    if (dfs(i) > k) {
      impossible();
    }
  priority_queue<pair<int, int>> Q;
  for (int i = 1; i <= n; i++) {
    v[dfs(i)].insert(i);
    if (in[i] == 0) {
      int val = a[i] ? -1 : dfs(i);
      Q.emplace(val, i);
    }
  }
  for (int i = k; i >= 1; i--) {
    if (v[i].size() == 0) {
      while (!Q.empty() && del[Q.top().second]) Q.pop();
      if (!Q.empty()) {
        int x = Q.top().second;
        int y = Q.top().first;
        if (y == -1) impossible();
        v[y].erase(x);
        v[i].insert(x);
        a[x] = i;
      } else
        impossible();
    }
    for (int j : v[i]) {
      del[j] = true;
      for (int x : g[j]) {
        in[x] -= 1;
        if (in[x] == 0) {
          int val = a[x] ? -1 : dfs(x);
          Q.emplace(val, x);
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int val = a[i] ? a[i] : dfs(i);
    cout << val << ' ';
  }
  cout << endl;
}