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 N = 200010;
vector<int> G[N];
vector<int> T[N];
bool cic;
int K[N];
int M[N];
int I[N];
int V[N];
int R[N];
int F[N];
int A[N];
void dfs(int x) {
  if (K[x] == 2) return;
  if (K[x] == 1) cic = true;
  if (cic) return;
  K[x] = 1;
  for (int i = 0; i < (int)G[x].size(); ++i) {
    int y = G[x][i];
    dfs(y);
  }
  K[x] = 2;
}
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &V[i]);
  for (int i = 0; i < m; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y);
    T[y].push_back(x);
    ++R[x];
    ++I[y];
  }
  queue<int> Q;
  cic = false;
  for (int i = 1; i <= n; ++i) {
    F[V[i]] = 1;
    M[i] = k;
    if (V[i] != 0) {
      Q.push(i);
      I[i] = 0;
      M[i] = V[i];
    } else if (I[i] == 0)
      Q.push(i);
    if (K[i] == 0) dfs(i);
  }
  if (cic) {
    printf("-1\n");
    return 0;
  }
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < (int)G[x].size(); ++i) {
      int y = G[x][i];
      --I[y];
      M[y] = min(M[y], M[x] - 1);
      if (I[y] == 0) Q.push(y);
    }
  }
  priority_queue<pair<int, int> > &qP = *new priority_queue<pair<int, int> >;
  for (int i = 1; i <= n; ++i) {
    if (R[i] == 0) qP.push({-M[i], i});
    if (M[i] < V[i] or M[i] <= 0) {
      printf("-1\n");
      return 0;
    }
  }
  set<int> &S = *new set<int>;
  for (int i = 1; i <= k; ++i) {
    if (F[i] == 0) S.insert(i);
  }
  for (int i = 0; i <= n; ++i) A[i] = 1;
  while (!qP.empty()) {
    int x = qP.top().second;
    qP.pop();
    if (V[x] == 0) {
      auto it = S.lower_bound(A[x]);
      if (!S.size())
        V[x] = M[x];
      else if (it != S.end() and *it <= M[x]) {
        V[x] = *it;
        S.erase(it);
      } else
        V[x] = A[x];
    }
    for (int i = 0; i < (int)T[x].size(); ++i) {
      int y = T[x][i];
      A[y] = max(A[y], V[x] + 1);
      --R[y];
      if (R[y] == 0) qP.push({-M[y], y});
    }
  }
  if (S.size()) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (i - 1) printf(" ");
    printf("%d", V[i]);
  }
  printf("\n");
}