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;
int n, m, k, a[N], deg[N], id[N], cnt, mi[N], mx[N], ans[N];
bool used[N];
vector<int> g[N], r[N];
vector<pair<int, int> > v[N];
queue<int> q;
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pq;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i])
      used[mx[i] = mi[i] = a[i]];
    else {
      mx[i] = k;
      mi[i] = 1;
    }
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    r[v].push_back(u);
    deg[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    id[++cnt] = u;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      if (!(--deg[v])) q.push(v);
    }
  }
  if (cnt < n) {
    cout << "-1\n";
    return 0;
  }
  for (int it = n, i; it; it--) {
    i = id[it];
    for (int j = 0; j < g[i].size(); j++) mi[i] = max(mi[i], mi[g[i][j]] + 1);
  }
  for (int it = 1, i; it <= n; it++) {
    i = id[it];
    for (int j = 0; j < r[i].size(); j++) mx[i] = min(mx[i], mx[r[i][j]] - 1);
  }
  for (int i = 1; i <= n; i++) {
    if (mi[i] > mx[i]) {
      cout << "-1\n";
      return 0;
    }
    v[mi[i]].push_back(pair<int, int>(mx[i], i));
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < v[i].size(); j++) pq.push(v[i][j]);
    if (!used[i]) {
      if (pq.empty()) {
        cout << "-1\n";
        return 0;
      }
      ans[pq.top().second] = i;
      pq.pop();
    }
    while (!pq.empty() && pq.top().first == i) {
      ans[pq.top().second] = i;
      pq.pop();
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}