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 = 200006;
struct Tnode {
  int low, top, i;
  bool operator<(const Tnode &t) const { return top > t.top; }
} a[MAXN];
int head[2][MAXN], cnt[2][MAXN];
int _next[MAXN * 2], node[MAXN * 2];
int line[MAXN];
int ans[MAXN];
int n, m, k, fp = 0;
bool _cmp(Tnode A, Tnode B) { return A.low < B.low; }
void _add(int x, int y, int k) {
  _next[fp] = head[k][x];
  node[fp] = y;
  head[k][x] = fp++;
}
bool _init(int t) {
  int he = 0, ta = 0, u, v;
  for (int i = 1; i <= n; i++)
    if (!cnt[t][i]) line[ta++] = i;
  while (he < ta) {
    u = line[he++];
    for (int i = head[t][u]; i != -1; i = _next[i]) {
      v = node[i];
      cnt[t][v]--;
      if (t)
        a[v].top = min(a[v].top, a[u].top - 1);
      else
        a[v].low = max(a[v].low, a[u].low + 1);
      if (!cnt[t][v]) line[ta++] = v;
    }
  }
  if (ta < n) return 1;
  return 0;
}
bool solve() {
  Tnode tmp;
  int j = 1;
  priority_queue<Tnode> q;
  bool flag;
  for (int i = 1; i <= k; i++) {
    flag = 0;
    for (; j <= n && a[j].low == i; j++) {
      tmp = a[j];
      tmp.i = j;
      q.push(tmp);
    }
    while (!q.empty() && q.top().top == i) {
      tmp = q.top();
      q.pop();
      ans[a[tmp.i].i] = i;
      flag = true;
    }
    if (!flag) {
      if (q.empty()) return 1;
      tmp = q.top();
      q.pop();
      ans[a[tmp.i].i] = i;
    }
  }
  return 0;
}
bool check() {
  for (int i = 1; i <= n; i++)
    if (a[i].top < a[i].low) return 1;
  return 0;
}
int main() {
  memset(head, -1, sizeof head);
  int x, y;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[i].i = i;
    if (x)
      a[i].top = x, a[i].low = x;
    else
      a[i].top = k, a[i].low = 1;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    _add(y, x, 0);
    _add(x, y, 1);
    cnt[0][x]++;
    cnt[1][y]++;
  }
  if (_init(0) || _init(1)) return 0 & printf("-1\n");
  sort(a + 1, a + n + 1, _cmp);
  if (check() || solve())
    printf("-1\n");
  else
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}