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;
void perdi() {
  puts("-1");
  exit(0);
}
vector<int> g[200005], gi[200005];
int r[200005];
int dp0[200005];
int dp1[200005];
int n, m, k;
priority_queue<pair<int, int> > q;
vector<int> w[200005];
int f0(int x) {
  if (dp0[x] >= 0) return dp0[x];
  if (dp0[x] == -2) perdi();
  if (r[x] >= 0) return dp0[x] = r[x];
  dp0[x] = -2;
  int rr = 0;
  for (int y : g[x]) rr = max(rr, f0(y) + 1);
  if (rr >= k) perdi();
  return dp0[x] = rr;
}
int f1(int x) {
  if (dp1[x] >= 0) return dp1[x];
  if (r[x] >= 0) return dp1[x] = r[x];
  int rr = k - 1;
  for (int y : gi[x]) rr = min(rr, f1(y) - 1);
  if (rr < 0) perdi();
  return dp1[x] = rr;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) scanf("%d", r + i), r[i]--;
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    g[x].push_back(y);
    gi[y].push_back(x);
  }
  memset(dp0, -1, sizeof(dp0));
  memset(dp1, -1, sizeof(dp1));
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) w[f0(i)].push_back(i);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) 0;
  for (int i = 0, ThxDem = k; i < ThxDem; ++i) {
    for (int x : w[i]) q.push(make_pair(-f1(x), x));
    while (!q.empty() && -q.top().first < i) q.pop();
    if (q.empty()) 0, perdi();
    r[q.top().second] = i;
    q.pop();
  }
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    if (r[i] < 0) r[i] = f1(i);
  for (int x = 0, ThxDem = n; x < ThxDem; ++x)
    for (int y : g[x])
      if (r[x] <= r[y]) perdi();
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) {
    if (i) putchar(' ');
    printf("%d", r[i] + 1);
  }
  puts("");
  return 0;
}