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 MOD = 1e9 + 7;
int n, m, k;
int a[200100];
vector<int> son[200100], par[200100];
int visit[200100], cnt[200100];
vector<int> order;
pair<int, int> l[200100];
vector<pair<pair<int, int>, int> > v;
int ans[200100];
void dfs(int a) {
  order.push_back(a);
  visit[a] = 1;
  for (int i = 0; i < son[a].size(); i++) {
    int u = son[a][i];
    if (visit[u]) {
      puts("-1");
      exit(0);
    }
    cnt[u]++;
    if (cnt[u] == par[u].size()) dfs(u);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    son[u].push_back(v);
    par[v].push_back(u);
  }
  int chk = 0;
  for (int i = 0; i < (int)(n); ++i)
    if (par[i].size() == 0) chk = 1;
  if (!chk) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (par[i].size() == 0) dfs(i);
  for (int i = 0; i < (int)(n); ++i) l[i] = make_pair(1, k);
  for (int i = 0; i < (int)(n); ++i)
    if (a[i] != 0) l[i] = make_pair(a[i], a[i]);
  for (int i = 0; i < n; i++) {
    int u = order[i];
    for (int j = 0; j < par[u].size(); j++) {
      int v = par[u][j];
      l[u].second = min(l[u].second, l[v].second - 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int u = order[i];
    for (int j = 0; j < son[u].size(); j++) {
      int v = son[u][j];
      l[u].first = max(l[u].first, l[v].first + 1);
    }
  }
  for (int i = 0; i < (int)(n); ++i)
    if (l[i].first > l[i].second) {
      puts("-1");
      return 0;
    }
  for (int i = 0; i < n; i++)
    v.push_back(make_pair(make_pair(l[i].first, l[i].second), i));
  sort(v.begin(), v.end());
  set<pair<int, int> > st;
  int t = 0;
  for (int i = 1; i <= k; i++) {
    while (t < n && v[t].first.first == i)
      st.insert(make_pair(v[t].first.second, v[t].second)), t++;
    if (!st.size()) {
      puts("-1");
      return 0;
    }
    int chk = 0;
    while (st.size()) {
      pair<int, int> tmp = *st.begin();
      if (tmp.first != i) break;
      chk = 1;
      ans[tmp.second] = i;
      st.erase(st.begin());
    }
    if (!chk) {
      pair<int, int> tmp = *st.begin();
      ans[tmp.second] = i;
      st.erase(st.begin());
    }
  }
  for (int i = 0; i < (int)(n); ++i) printf("%d ", ans[i]);
  return 0;
}