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;
int a[202000], b[202000], l[202000], r[202000], used[202000];
vector<int> G[202000];
int deg[202000], q[202000], he, ta;
struct P {
  int id, st, en;
} p[202000];
bool cp(P p1, P p2) { return p1.st < p2.st; }
struct cmp {
  bool operator()(P p1, P p2) { return p1.en > p2.en; }
};
priority_queue<P, vector<P>, cmp> pq;
int main() {
  scanf("%d%d%d", &n, &m, &k);
  if (k > n) return 0 * puts("-1");
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    used[a[i]] = 1;
    b[i] = a[i];
    if (b[i]) {
      l[i] = r[i] = a[i];
    } else {
      l[i] = 1;
      r[i] = k;
    }
  }
  int x, y;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    deg[y]++;
  }
  he = ta = 0;
  for (int i = 1; i <= n; i++) {
    if (!deg[i]) q[++ta] = i;
  }
  int now;
  for (; he < ta;) {
    now = q[++he];
    for (int i = 0; i < (int)G[now].size(); i++) {
      y = G[now][i];
      if (!--deg[y]) q[++ta] = y;
    }
  }
  if (ta < n) return 0 * printf("-1");
  for (int j = 1; j <= n; j++) {
    now = q[j];
    for (int i = 0; i < (int)G[now].size(); i++) {
      y = G[now][i];
      r[y] = min(r[y], r[now] - 1);
    }
  }
  for (int j = n; j > 0; j--) {
    now = q[j];
    for (int i = 0; i < (int)G[now].size(); i++) {
      y = G[now][i];
      l[now] = max(l[now], l[y] + 1);
      if (l[now] > r[now]) return 0 * printf("-1");
    }
  }
  for (int i = 1; i <= n; i++) {
    p[i] = (P){i, l[i], r[i]};
  }
  sort(p + 1, p + n + 1, cp);
  int pos = 0, fl;
  for (int i = 1; i <= k; i++) {
    while (pos < n && p[pos + 1].st <= i) pq.push(p[++pos]);
    fl = used[i];
    for (; !pq.empty();) {
      if (pq.top().en == i || !fl) {
        a[pq.top().id] = i;
        pq.pop();
      } else {
        break;
      }
      fl = 1;
    }
    if (!fl) return 0 * printf("-1");
  }
  for (int j = 1; j <= n; j++) {
    now = q[j];
    for (int i = 0; i < (int)G[now].size(); i++) {
      y = G[now][i];
      a[y] = min(a[y], a[now] - 1);
    }
    if (b[now] && a[now] != b[now]) return 0 * printf("-1");
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
}