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;
template <class T>
bool cmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool cmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const int N = 2e5 + 10;
int n, m, nranks;
vector<int> E[N];
int deg[N];
int q[N], qh, qt;
int l[N], r[N];
int cnt[N];
vector<int> nodes[N];
void lose() {
  puts("-1");
  exit(EXIT_SUCCESS);
}
int main() {
  scanf("%d%d%d", &n, &m, &nranks);
  for (int i = 1; i <= n; i++) scanf("%d", l + i);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 0) q[qt++] = i;
    if (l[i] == 0)
      l[i] = 1, r[i] = nranks;
    else
      cnt[r[i] = l[i]]++;
  }
  while (qh < qt) {
    int u = q[qh++];
    for (int v : E[u])
      if (--deg[v] == 0) q[qt++] = v;
  }
  if (qt != n) {
    lose();
  }
  for (int qi = 0; qi < n; qi++) {
    {
      int u = q[qi];
      for (int v : E[u]) cmin(r[v], r[u] - 1);
    }
    {
      int u = q[n - qi - 1];
      for (int v : E[u]) cmax(l[u], l[v] + 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (l[i] > r[i]) lose();
    nodes[l[i]].push_back(i);
  }
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  for (int rnk = 1; rnk <= nranks; rnk++) {
    for (int u : nodes[rnk]) pq.push({r[u], u});
    if (cnt[rnk] == 0) {
      if (pq.empty()) lose();
      l[pq.top().second] = rnk;
      pq.pop();
    }
    while (!pq.empty() && pq.top().first == rnk) {
      l[pq.top().second] = rnk;
      pq.pop();
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", l[i], " \n"[i == n]);
  for (int i = 1; i <= n; i++) {
    for (int v : E[i]) {
      if (l[i] <= l[v]) {
        printf("(%d, %d)\n", i, v);
      }
      assert(l[i] > l[v]);
    }
  }
  return 0;
}