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, w[201000], Q[201000], head, tail, Deg[201000], Mn[201000],
    Mx[201000], Res[201000];
vector<int> E[201000], F[201000];
set<int> Set;
struct point {
  int b, e, num;
  bool operator<(const point &p) const { return e != p.e ? e < p.e : b < p.b; }
} P[201000];
int main() {
  int i, a, b, j;
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    E[b].push_back(a);
    F[a].push_back(b);
  }
  for (i = 1; i <= n; i++) {
    Deg[i] = F[i].size();
    if (!Deg[i]) Q[++tail] = i;
  }
  for (i = 1; i <= n; i++) Mn[i] = 1, Mx[i] = K;
  while (head < tail) {
    int x = Q[++head];
    if (w[x]) Mn[x] = max(Mn[x], w[x]);
    for (i = 0; i < E[x].size(); i++) {
      Deg[E[x][i]]--;
      Mn[E[x][i]] = max(Mn[E[x][i]], Mn[x] + 1);
      if (!Deg[E[x][i]]) Q[++tail] = E[x][i];
    }
  }
  if (tail != n) {
    printf("-1\n");
    return 0;
  }
  head = tail = 0;
  for (i = 1; i <= n; i++) {
    Deg[i] = E[i].size();
    if (!Deg[i]) Q[++tail] = i;
  }
  while (head < tail) {
    int x = Q[++head];
    if (w[x]) Mx[x] = min(Mx[x], w[x]);
    for (i = 0; i < F[x].size(); i++) {
      Deg[F[x][i]]--;
      Mx[F[x][i]] = min(Mx[F[x][i]], Mx[x] - 1);
      if (!Deg[F[x][i]]) Q[++tail] = F[x][i];
    }
  }
  for (i = 1; i <= n; i++) {
    if (Mn[i] > Mx[i]) {
      printf("-1\n");
      return 0;
    }
    P[i] = {Mn[i], Mx[i], i};
  }
  sort(P + 1, P + n + 1);
  for (i = 1; i <= K; i++) Set.insert(i);
  for (i = 1; i <= n; i++) {
    int x = P[i].num;
    auto it = Set.lower_bound(Mn[x]);
    if (it == Set.end() || (*it) > Mx[x]) {
      Res[x] = Mn[x];
    } else {
      Res[x] = *it;
      Set.erase(it);
    }
    for (j = 0; j < E[x].size(); j++) {
      Mn[E[x][j]] = max(Mn[E[x][j]], Res[x] + 1);
    }
  }
  if (Set.empty()) {
    for (i = 1; i <= n; i++) printf("%d ", Res[i]);
  } else
    printf("-1\n");
}