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 = 2e5 + 5;
int n, m, k, r[N], mn[N], mx[N], deg[N], ans[N], s[N];
vector<int> e[N], re[N], lvl[N];
set<pair<int, int> > cur;
void solve() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    mn[i] = r[i] ? r[i] : 1;
    mx[i] = r[i] ? r[i] : k;
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    re[v].push_back(u);
  }
  int hd = 0, tl = 0;
  for (int i = 1; i <= n; i++) {
    deg[i] = re[i].size();
    if (!deg[i]) s[tl++] = i;
  }
  while (hd < tl) {
    for (auto v : e[s[hd]]) {
      mx[v] = min(mx[v], mx[s[hd]] - 1);
      deg[v]--;
      if (!deg[v]) s[tl++] = v;
    }
    hd++;
  }
  if (tl != n) {
    puts("-1");
    return;
  }
  hd = tl = 0;
  for (int i = 1; i <= n; i++) {
    deg[i] = e[i].size();
    if (!deg[i]) s[tl++] = i;
  }
  while (hd < tl) {
    for (auto v : re[s[hd]]) {
      mn[v] = max(mn[v], mn[s[hd]] + 1);
      deg[v]--;
      if (!deg[v]) s[tl++] = v;
    }
    hd++;
  }
  if (tl != n) {
    puts("-1");
    return;
  }
  for (int i = 1; i <= n; i++)
    if (mx[i] < mn[i]) {
      puts("-1");
      return;
    }
  for (int i = 1; i <= n; i++) {
    lvl[mn[i]].push_back(i);
  }
  for (int i = 1; i <= k; i++) {
    for (auto it : lvl[i]) cur.insert(make_pair(mx[it], it));
    if (cur.empty()) {
      puts("-1");
      return;
    }
    ans[cur.begin()->second] = i;
    cur.erase(cur.begin());
    while (!cur.empty() && cur.begin()->first == i) {
      ans[cur.begin()->second] = i;
      cur.erase(cur.begin());
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  solve();
  return 0;
}