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;
int n, m, k;
int used[200200];
vector<int> g[200100];
vector<int> gg[200100];
int l[200200];
int r[200200];
vector<int> ord;
void dfs(int v) {
  used[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) dfs(to);
  }
  ord.push_back(v);
}
vector<int> op[200200];
vector<int> cl[200200];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    if (x == 0) {
      l[i] = 1;
      r[i] = k;
    } else {
      l[i] = x;
      r[i] = x;
    }
  }
  for (int i = 0, a, b; i < m; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    gg[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      dfs(i);
    }
  }
  reverse(ord.begin(), ord.end());
  for (int i = 0; i < n; i++) {
    int v = ord[i];
    for (int j = 0; j < g[v].size(); j++) {
      int to = g[v][j];
      r[to] = min(r[to], r[v] - 1);
    }
  }
  reverse(ord.begin(), ord.end());
  for (int i = 0; i < n; i++) {
    int v = ord[i];
    for (int j = 0; j < gg[v].size(); j++) {
      int to = gg[v][j];
      l[to] = max(l[to], l[v] + 1);
    }
  }
  vector<pair<pair<int, int>, int> > cur;
  for (int i = 1; i <= n; i++) {
    if (l[i] > r[i]) {
      printf("-1\n");
      return 0;
    }
    op[l[i]].push_back(i);
    cl[r[i]].push_back(i);
  }
  int sz = 0;
  set<pair<int, int> > S;
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < op[i].size(); j++) {
      int t = op[i][j];
      S.insert(make_pair(r[t], t));
    }
    if (S.size() == 0) {
      cout << -1 << endl;
      return 0;
    }
    r[S.begin()->second] = i;
    S.erase(S.begin());
    for (int j = 0; j < cl[i].size(); j++) {
      int t = cl[i][j];
      S.erase(make_pair(r[t], t));
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%d", r[i]);
  }
  printf("\n");
  return 0;
}