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 = 1e6 + 100;
int r[N], hi[N], vis[N], lo[N];
vector<int> sorted;
vector<int> G[N], rG[N];
void dfs(int u) {
  vis[u] = 1;
  for (int v : G[u]) {
    if (vis[v] == 2) continue;
    if (vis[v] == 1) {
      cout << "-1" << endl;
      exit(0);
    }
    dfs(v);
  }
  vis[u] = 2;
  sorted.push_back(u);
}
bool cmp(int u, int v) { return hi[u] < hi[v]; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, K;
  cin >> n >> m >> K;
  for (int i = 1; i <= n; i++) {
    cin >> r[i];
    hi[i] = (r[i] ? r[i] : K);
    lo[i] = (r[i] ? r[i] : 1);
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> y >> x;
    G[x].push_back(y);
    rG[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  for (int u : sorted) {
    for (int v : G[u]) {
      hi[u] = min(hi[u], hi[v] - 1);
    }
  }
  sort(sorted.begin(), sorted.end(), cmp);
  set<int> baki;
  for (int i = 1; i <= K; i++) baki.insert(i);
  for (int u : sorted) {
    for (int v : rG[u]) {
      lo[u] = max(lo[u], lo[v] + 1);
    }
    if (lo[u] > hi[u]) {
      cout << "-1" << endl;
      exit(0);
    }
    auto it = baki.lower_bound(lo[u]);
    if (it == baki.end() or (*it) > hi[u]) {
    } else {
      lo[u] = *it;
      baki.erase(lo[u]);
    }
  }
  if (baki.size() > 0) {
    cout << "-1" << endl;
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    cout << lo[i] << " ";
  }
  return 0;
}