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;
inline int Read() {
  int x = 0, f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 200005;
int n, m, k, ql, qr, q[MAXN], d[MAXN], l[MAXN], r[MAXN], ans[MAXN];
vector<int> adj[MAXN], add[MAXN];
set<pair<int, int> > cur;
int main() {
  n = Read(), m = Read(), k = Read();
  for (int i = 1; i <= n; i++) {
    int x = Read();
    if (!x)
      l[i] = 1, r[i] = k;
    else
      l[i] = r[i] = x;
  }
  for (int i = 1, x, y; i <= m; i++)
    x = Read(), y = Read(), adj[x].push_back(y), d[y]++;
  for (int i = 1; i <= n; i++)
    if (!d[i]) q[++qr] = i;
  while (ql ^ qr) {
    int x = q[++ql];
    for (auto y : adj[x])
      if (!--d[y]) q[++qr] = y;
  }
  if (qr ^ n) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) {
    int x = q[i];
    for (auto y : adj[x]) r[y] = min(r[y], r[x] - 1);
  }
  for (int i = n; i; i--) {
    int x = q[i];
    for (auto y : adj[x]) l[x] = max(l[x], l[y] + 1);
  }
  for (int i = 1; i <= n; i++)
    if (l[i] > r[i]) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) add[l[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (auto x : add[i]) cur.insert(make_pair(r[x], x));
    if (cur.empty()) return puts("-1"), 0;
    pair<int, int> v = *cur.begin();
    ans[v.second] = i;
    cur.erase(cur.begin());
    while (!cur.empty() && (v = *cur.begin()).first == i)
      ans[v.second] = i, cur.erase(cur.begin());
  }
  for (int x = 1; x <= n; x++)
    for (auto y : adj[x])
      if (ans[x] <= ans[y]) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}