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>
#pragma GCC optimize(2)
using namespace std;
void qread(int &x) {
  int neg = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
const int maxn = 200005;
int n, m, k;
int deg[maxn], q[maxn], rk[maxn];
int low[maxn], upp[maxn], ord[maxn];
vector<int> edge[maxn], rev[maxn];
set<pair<int, int> > s;
bool cmp(int a, int b) { return low[a] < low[b]; }
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int(i) = 0; (i) < n; i++) {
    low[i] = 1;
    upp[i] = k;
    scanf("%d", &rk[i]);
    if (rk[i]) low[i] = upp[i] = rk[i];
  }
  for (int(i) = 0; (i) < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    deg[v]++;
    edge[u].push_back(v);
    rev[v].push_back(u);
  }
  int l = 0, r = 0;
  for (int(i) = 0; (i) < n; i++)
    if (!deg[i]) q[r++] = i;
  while (l < r) {
    int x = q[l++];
    for (int(i) = 0; (i) < edge[x].size(); i++) {
      deg[edge[x][i]]--;
      if (!deg[edge[x][i]]) q[r++] = edge[x][i];
    }
  }
  if (r != n) {
    puts("-1");
    return 0;
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = q[i];
    for (int(j) = 0; (j) < edge[x].size(); j++) {
      low[x] = max(low[x], low[edge[x][j]] + 1);
    }
  }
  for (int(i) = 0; (i) < n; i++) {
    int x = q[i];
    for (int(j) = 0; (j) < rev[x].size(); j++) {
      upp[x] = min(upp[x], upp[rev[x][j]] - 1);
    }
  }
  for (int(i) = 0; (i) < n; i++) {
    ord[i] = i;
    if (low[i] > upp[i]) {
      puts("-1");
      return 0;
    }
  }
  int ptr = 0;
  sort(ord, ord + n, cmp);
  for (int(i) = 1; (i) <= k; i++) {
    while (ptr < n && low[ord[ptr]] <= i) {
      s.insert(make_pair(upp[ord[ptr]], ord[ptr]));
      ptr++;
    }
    bool ok = false;
    while (!s.empty()) {
      pair<int, int> x = *s.begin();
      if (x.first != i) break;
      ok = true;
      rk[x.second] = i;
      s.erase(s.begin());
    }
    if (!ok) {
      if (s.empty()) {
        puts("-1");
        return 0;
      }
      pair<int, int> x = *s.begin();
      rk[x.second] = i;
      s.erase(s.begin());
    }
  }
  for (int(i) = 0; (i) < n; i++) printf("%d ", rk[i]);
  puts("");
  return 0;
}