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;
int n, m, k, l[200010], r[200010], v[200010], ans[200010];
class graph {
 public:
  vector<int> G[200010];
  int deg[200010];
  void addedge(int s, int e) {
    G[s].push_back(e);
    deg[e]++;
  }
} G1, G2;
vector<int> P[200010];
priority_queue<pair<int, int> > Q;
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    if (v[i])
      l[i] = r[i] = v[i];
    else
      l[i] = 1, r[i] = k;
  }
  for (int i = 1, ti, tj; i <= m; i++) {
    scanf("%d%d", &ti, &tj);
    G1.addedge(ti, tj);
    G2.addedge(tj, ti);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (G1.deg[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int fr = q.front();
    q.pop();
    for (int x : G1.G[fr]) {
      r[x] = min(r[x], r[fr] - 1);
      G1.deg[x]--;
      if (!G1.deg[x]) q.push(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (G1.deg[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (G2.deg[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int fr = q.front();
    q.pop();
    for (int x : G2.G[fr]) {
      l[x] = max(l[x], l[fr] + 1);
      G2.deg[x]--;
      if (!G2.deg[x]) q.push(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (G2.deg[i] || l[i] > r[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) P[r[i]].push_back(i);
  for (int i = k; i >= 1; i--) {
    for (int j : P[i]) Q.push(make_pair(l[j], j));
    if (Q.empty()) {
      puts("-1");
      return 0;
    }
    int tp = Q.top().second;
    ans[tp] = i;
    Q.pop();
    while (!Q.empty()) {
      tp = Q.top().second;
      if (l[tp] < i) break;
      Q.pop();
      ans[tp] = i;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}