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 long long MAXN = 2e5 + 10;
set<pair<long long, long long> > s;
vector<pair<long long, long long> > e[MAXN];
vector<long long> vertex[MAXN];
vector<long long> ts;
long long a[MAXN];
long long dpl[MAXN];
long long dpr[MAXN];
bool check[MAXN];
void topological_sort(long long v) {
  check[v] = true;
  for (long long i = 0; i < vertex[v].size(); i++) {
    if (!check[vertex[v][i]]) {
      topological_sort(vertex[v][i]);
    }
  }
  ts.push_back(v);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < m; i++) {
    long long v, u;
    cin >> v >> u;
    vertex[v].push_back(u);
  }
  for (long long i = 1; i <= n; i++) {
    if (!check[i]) {
      topological_sort(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] > 0) {
      dpl[i] = a[i];
      dpr[i] = a[i];
    } else {
      dpl[i] = 1;
      dpr[i] = k;
    }
  }
  for (long long i = 0; i < ts.size(); i++) {
    long long v = ts[i];
    for (long long j = 0; j < vertex[v].size(); j++) {
      dpl[v] = max(dpl[v], dpl[vertex[v][j]] + 1);
    }
  }
  for (long long i = (long long)(ts.size()) - 1; i >= 0; i--) {
    long long v = ts[i];
    for (long long j = 0; j < vertex[v].size(); j++) {
      dpr[vertex[v][j]] = min(dpr[vertex[v][j]], dpr[v] - 1);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (a[i] > 0 && (dpl[i] != a[i] || dpr[i] != a[i])) {
      cout << -1;
      return 0;
    }
    if (dpr[i] < dpl[i]) {
      cout << -1;
      return 0;
    }
    e[dpl[i]].push_back(make_pair(dpr[i], i));
  }
  for (long long i = 1; i <= k; i++) {
    for (long long j = 0; j < e[i].size(); j++) {
      s.insert(e[i][j]);
    }
    while (s.size() > 0 && (s.begin()->first) < i) {
      s.erase(s.begin());
    }
    if (s.size() == 0) {
      cout << -1;
      return 0;
    }
    long long v = (s.begin()->second);
    s.erase(s.begin());
    a[v] = i;
  }
  for (long long i = 0; i < ts.size(); i++) {
    long long v = ts[i];
    if (a[v] == 0) {
      a[v] = 1;
      for (long long j = 0; j < vertex[v].size(); j++) {
        a[v] = max(a[v], a[vertex[v][j]] + 1);
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
}