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;
void test_case() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> out(n), p(n);
  vector<vector<int> > g1(n), g2(n);
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g1[a].push_back(b);
    g2[b].push_back(a);
    out[a]++;
  }
  vector<int> topologicOrder;
  {
    vector<int> vis(n);
    function<void(int)> dfs = [&](int u) {
      if (vis[u] == 2) {
        cout << -1 << endl;
        exit(0);
      }
      if (vis[u]) return;
      vis[u] = 2;
      for (auto v : g1[u]) {
        dfs(v);
      }
      topologicOrder.push_back(u);
      vis[u] = 1;
    };
    for (int i = 0; i < n; i++) {
      dfs(i);
    }
  }
  vector<int> min_v(n, 1), max_v(n, k);
  for (int i = 0; i < n; i++)
    if (p[i]) min_v[i] = max_v[i] = p[i];
  for (int i = 0; i < n; i++) {
    int u = topologicOrder[i];
    for (auto v : g2[u]) min_v[v] = max(min_v[v], min_v[u] + 1);
  }
  for (int i = n - 1; i >= 0; i--) {
    int u = topologicOrder[i];
    for (auto v : g1[u]) max_v[v] = min(max_v[v], max_v[u] - 1);
  }
  priority_queue<pair<int, int> > q, events;
  for (int i = 0; i < n; i++) {
    if (min_v[i] > max_v[i] || min_v[i] > k || max_v[i] < 1) {
      cout << -1 << endl;
      exit(0);
    }
    if (!out[i]) {
      events.push({-min_v[i], i});
    }
  }
  function<void(int)> getNext = [&](int u) {
    for (auto v : g2[u]) {
      if (out[v]-- == 1) {
        events.push({-min_v[v], v});
      }
    }
  };
  function<void(int)> prop = [&](int i) {
    while (!events.empty() && -events.top().first <= i) {
      int x = events.top().second;
      events.pop();
      q.push({-max_v[x], x});
    }
  };
  for (int i = 1; i <= k; i++) {
    prop(i);
    if (q.empty()) {
      cout << -1 << endl;
      exit(0);
    }
    int x = q.top().second;
    if (min_v[x] > i || max_v[x] < i) {
      cout << -1 << endl;
      exit(0);
    }
    p[x] = i;
    getNext(x);
    prop(i);
    q.pop();
    while (!q.empty() && -q.top().first == i) {
      x = q.top().second;
      q.pop();
      p[x] = i;
      getNext(x);
      prop(i);
    }
  }
  for (int i = 0; i < n; i++) {
    cout << p[i] << " ";
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}