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;
int rk[N], down[N], up[N], a[N];
int in[N], out[N];
int q[N], h, t;
int n, m, K;
vector<int> vec[N];
vector<int> vi[N], vo[N];
struct data {
  int w, x;
  data() {}
  data(int w, int x) : w(w), x(x) {}
  friend bool operator<(data aa, data bb) { return aa.w > bb.w; }
};
priority_queue<data> minh;
void work(int x, int now) {
  int i, y;
  rk[x] = now;
  for (i = 0; i <= (int)vi[x].size() - 1; i++) {
    y = vi[x][i];
    down[y] = max(down[y], rk[x] + 1);
    in[y]--;
    if (!in[y]) vec[down[y]].push_back(y);
  }
}
int main() {
  int i, x, y, j;
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++)
    if (a[i])
      up[i] = a[i];
    else
      up[i] = K;
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    swap(x, y);
    out[x]++;
    vo[y].push_back(x);
    in[y]++;
    vi[x].push_back(y);
  }
  h = t = 0;
  for (i = 1; i <= n; i++)
    if (out[i] == 0) q[++t] = i;
  while (h < t) {
    x = q[++h];
    for (i = 0; i <= (int)vo[x].size() - 1; i++) {
      y = vo[x][i];
      out[y]--;
      up[y] = min(up[y], up[x] - 1);
      if (!out[y]) q[++t] = y;
    }
  }
  for (i = 1; i <= n; i++)
    if (a[i] && up[i] < a[i]) {
      printf("-1\n");
      return 0;
    }
  for (i = 1; i <= n; i++)
    if (up[i] < 1) {
      printf("-1\n");
      return 0;
    }
  for (i = 1; i <= n; i++)
    if (a[i])
      down[i] = a[i];
    else
      down[i] = 1;
  for (i = 1; i <= n; i++)
    if (!in[i]) vec[down[i]].push_back(i);
  for (i = 1; i <= K; i++) {
    for (j = 0; j <= (int)vec[i].size() - 1; j++) {
      x = vec[i][j];
      minh.push(data(up[x], x));
    }
    if (minh.empty()) {
      printf("-1\n");
      return 0;
    }
    work(minh.top().x, i);
    minh.pop();
    while (!minh.empty() && minh.top().w == i) {
      work(minh.top().x, i);
      minh.pop();
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", rk[i]);
}