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 M = 200100;
int n, m, k, a[M], l[M], r[M], x, y, du[M], p[M], cnt, ans[M];
vector<int> g[M], add[M];
queue<int> q;
set<pair<int, int> > st;
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i])
      l[i] = r[i] = a[i];
    else
      l[i] = 1, r[i] = k;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    du[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (!du[i]) q.push(i);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    p[++cnt] = now;
    for (int i = 0; i < g[now].size(); i++) {
      int v = g[now][i];
      du[v]--;
      if (!du[v]) q.push(v);
    }
  }
  if (cnt != n) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < g[p[i]].size(); j++) {
      int v = g[p[i]][j];
      r[v] = min(r[v], r[p[i]] - 1);
    }
  for (int i = n; i >= 1; i--)
    for (int j = 0; j < g[p[i]].size(); j++) {
      int v = g[p[i]][j];
      l[p[i]] = max(l[v] + 1, l[p[i]]);
    }
  for (int i = 1; i <= n; i++)
    if (l[i] > r[i]) {
      puts("-1");
      return 0;
    } else
      add[l[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < add[i].size(); j++)
      st.insert(make_pair(r[add[i][j]], add[i][j]));
    if (st.empty()) {
      puts("-1");
      return 0;
    }
    pair<int, int> now = *st.begin();
    ans[now.second] = i;
    st.erase(st.begin());
    while (!st.empty()) {
      pair<int, int> now = *st.begin();
      if (now.first != i && i != k) break;
      ans[now.second] = i;
      st.erase(st.begin());
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < g[i].size(); j++)
      if (ans[i] < ans[g[i][j]]) {
        puts("-1");
        return 0;
      }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}