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 = 2e5 + 10;
int n, m, k;
int ans[MAXN], top[MAXN], bot[MAXN], a[MAXN];
int in[MAXN], ot[MAXN];
vector<int> vec[MAXN], vin[MAXN], vot[MAXN];
queue<int> q;
struct Tpeo {
  int tp, id;
  Tpeo() {}
  Tpeo(int tpp, int idd) {
    tp = tpp;
    id = idd;
  }
  friend bool operator<(Tpeo A, Tpeo B) { return A.tp > B.tp; }
};
priority_queue<Tpeo> mi;
void solve_bot(int u, int now) {
  ans[u] = now;
  for (int i = 0; i < vin[u].size(); ++i) {
    int v = vin[u][i];
    bot[v] = max(bot[v], ans[u] + 1);
    --in[v];
    if (!in[v]) vec[bot[v]].push_back(v);
  }
}
void init() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i])
      top[i] = a[i], bot[i] = a[i];
    else
      top[i] = k, bot[i] = 1;
  }
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    ot[v]++;
    vot[u].push_back(v);
    in[u]++;
    vin[v].push_back(u);
  }
}
void solve_top() {
  for (int i = 1; i <= n; ++i)
    if (!ot[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < vot[u].size(); ++i) {
      int v = vot[u][i];
      ot[v]--;
      top[v] = min(top[v], top[u] - 1);
      if (!ot[v]) q.push(v);
    }
  }
}
void solve() {
  solve_top();
  for (int i = 1; i <= n; ++i)
    if ((a[i] && top[i] < a[i]) || top[i] < 1) {
      printf("-1\n");
      return;
    }
  for (int i = 1; i <= n; ++i)
    if (!in[i]) vec[bot[i]].push_back(i);
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < vec[i].size(); ++j) {
      int x = vec[i][j];
      mi.push(Tpeo(top[x], x));
    }
    if (mi.empty()) {
      printf("-1\n");
      return;
    }
    solve_bot(mi.top().id, i);
    mi.pop();
    while (!mi.empty() && mi.top().tp == i) {
      solve_bot(mi.top().id, i);
      mi.pop();
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
}
int main() {
  init();
  solve();
  return 0;
}