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;
queue<int> q;
struct edge {
  int x, y, next;
} e[200010], a[200010];
int len = 0, last[200010];
int n, m, k, c[200010], d[200010], ans[200010];
struct node {
  int l, r;
} p[200010];
vector<int> vec[200010];
void ins(int x, int y) {
  a[++len].y = y;
  a[len].next = last[x];
  last[x] = len;
}
priority_queue<pair<int, int> > q1;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    if (c[i])
      p[i].l = p[i].r = c[i];
    else
      p[i].l = 1, p[i].r = k;
  }
  for (int i = 1; i <= m; i++) scanf("%d %d", &e[i].x, &e[i].y);
  len = 0;
  memset(last, 0, sizeof(last));
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= m; i++) {
    ins(e[i].x, e[i].y);
    d[e[i].y]++;
  }
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = last[x]; i; i = a[i].next) {
      int y = a[i].y;
      d[y]--;
      if (!d[y]) q.push(y);
      p[y].r = min(p[y].r, p[x].r - 1);
    }
  }
  len = 0;
  memset(last, 0, sizeof(last));
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= m; i++) {
    ins(e[i].y, e[i].x);
    d[e[i].x]++;
  }
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = last[x]; i; i = a[i].next) {
      int y = a[i].y;
      d[y]--;
      if (!d[y]) q.push(y);
      p[y].l = max(p[y].l, p[x].l + 1);
    }
  }
  for (int i = 1; i <= n; i++)
    if (d[i] || p[i].l > p[i].r) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) vec[p[i].r].push_back(i);
  for (int i = k; i >= 1; i--) {
    for (int j = 0; j < vec[i].size(); j++)
      q1.push(make_pair(p[vec[i][j]].l, vec[i][j]));
    if (q1.empty()) return puts("-1"), 0;
    int x = q1.top().second;
    q1.pop();
    ans[x] = i;
    while (!q1.empty()) {
      int x = q1.top().second;
      if (p[x].l < i) break;
      q1.pop();
      ans[x] = i;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}