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 maxn = 200005;
int n, m, K, lab[maxn], deg[maxn], Deg[maxn], r[maxn], L[maxn], R[maxn];
vector<int> rG[maxn], G[maxn], cand[maxn];
queue<int> q;
priority_queue<pair<int, int> > Q;
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &r[i]);
    L[i] = (r[i] == 0) ? 1 : r[i];
    R[i] = (r[i] == 0) ? K : r[i];
  }
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    rG[v].push_back(u);
    ++deg[v];
    ++Deg[u];
  }
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : G[u]) {
      R[v] = min(R[v], R[u] - 1);
      --deg[v];
      if (deg[v] == 0) q.push(v);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!Deg[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : rG[u]) {
      L[v] = max(L[v], L[u] + 1);
      --Deg[v];
      if (Deg[v] == 0) q.push(v);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (L[i] > R[i] || deg[i] > 0 || Deg[i] > 0) return 0 * puts("-1");
    deg[i] = rG[i].size();
    if (deg[i] == 0) cand[R[i]].push_back(i);
  }
  for (auto v : cand[K]) Q.push(pair<int, int>(L[v], v));
  int i = K;
  while (!Q.empty()) {
    bool ok = false;
    while (!Q.empty() && Q.top().first == i) {
      ok = true;
      int u = Q.top().second;
      Q.pop();
      lab[u] = i;
      for (auto v : G[u]) {
        --deg[v];
        if (deg[v] == 0) cand[min(R[v], i - 1)].push_back(v);
      }
    }
    if (!ok) {
      int u = Q.top().second;
      Q.pop();
      lab[u] = i;
      for (auto v : G[u]) {
        --deg[v];
        if (deg[v] == 0) cand[min(R[v], i - 1)].push_back(v);
      }
    }
    --i;
    for (auto v : cand[i]) Q.push(pair<int, int>(L[v], v));
  }
  if (i != 0) return 0 * puts("-1");
  for (int i = 1; i <= n; ++i) printf("%d%c", lab[i], i == n ? '\n' : ' ');
  return 0;
}