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;
template <class T>
void chkmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chkmin(T &a, T b) {
  if (a > b) a = b;
}
int n, m, up;
int a[200005], b[200005], bs;
int head[200005], to[200005], nxt[200005];
int mn[200005], mx[200005];
int G, vis[200005], tot;
void add_edge(int first, int second) {
  tot++;
  to[tot] = second;
  nxt[tot] = head[first];
  head[first] = tot;
  return;
}
bool dfs(int k) {
  vis[k] = 1;
  int first;
  for (int i = head[k]; i; i = nxt[i]) {
    first = to[i];
    if (vis[first] == 2)
      continue;
    else if (vis[first] || dfs(first))
      return 1;
  }
  vis[k] = 2;
  b[++bs] = k;
  return 0;
}
vector<int> v[200005];
int main() {
  int first, second;
  scanf("%d %d %d", &n, &m, &up);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &first, &second);
    add_edge(first, second);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i] && dfs(i)) {
      puts("-1");
      return 0;
    }
  bool flag = 0;
  for (int j = 1; j <= bs && !flag; j++) {
    int k = b[j];
    mn[k] = 1;
    for (int i = head[k]; i; i = nxt[i]) chkmax(mn[k], mn[to[i]] + 1);
    if (a[k] && mn[k] > a[k])
      flag = 1;
    else if (a[k])
      mn[k] = a[k];
  }
  for (int i = 1; i <= n; i++) mx[i] = up;
  for (int j = bs; j && !flag; j--) {
    int k = b[j];
    if (a[k] && mx[k] < a[k])
      flag = 1;
    else if (a[k])
      mx[k] = a[k];
    for (int i = head[k]; i; i = nxt[i]) chkmin(mx[to[i]], mx[k] - 1);
  }
  for (int i = 1; i <= n; i++)
    if (mn[i] > mx[i]) flag = 1;
  if (up > n) flag = 1;
  if (flag)
    puts("-1");
  else {
    for (int i = 1; i <= n; i++) v[mn[i]].push_back(i);
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        pq;
    for (int i = 1; i <= up; i++) {
      for (int j = 0; j < v[i].size(); j++)
        pq.push(pair<int, int>(mx[v[i][j]], v[i][j]));
      if (pq.empty()) {
        puts("-1");
        return 0;
      }
      pair<int, int> rpt = pq.top();
      pq.pop();
      a[rpt.second] = i;
      while (!pq.empty()) {
        rpt = pq.top();
        if (rpt.first == i)
          a[rpt.second] = i, pq.pop();
        else
          break;
      }
    }
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  }
  return 0;
}