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, x, rk[200005];
vector<int> G[200005];
vector<pair<int, int> > u[200005];
int color[200005], l[200005], r[200005];
bool used[200005];
deque<int> order;
set<pair<int, int> > s;
bool dfs_visit(int v) {
  color[v] = 1;
  bool f = true;
  for (int i = 0; i < G[v].size(); i++) {
    int to = G[v][i];
    if (color[to] == 0) f |= dfs_visit(to);
    if (color[to] == 1) return false;
  }
  color[v] = 2;
  order.push_front(v);
  return f;
}
bool toposort() {
  memset(color, 0, sizeof(color));
  for (int i = 1; i <= n; i++)
    if (color[i] == 0)
      if (!dfs_visit(i)) return false;
  return true;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x != 0)
      l[i] = r[i] = x;
    else {
      r[i] = k;
      l[i] = 1;
    }
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
  }
  if (!toposort()) {
    puts("-1");
    return 0;
  }
  memset(used, false, sizeof(used));
  for (int i = 0; i < n; i++) {
    int v = order[i];
    for (int j = 0; j < G[v].size(); j++) {
      int to = G[v][j];
      r[to] = min(r[to], r[v] - 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int v = order[i];
    for (int j = 0; j < G[v].size(); j++) {
      int to = G[v][j];
      l[v] = max(l[to] + 1, l[v]);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (l[i] > r[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) u[l[i]].push_back(pair<int, int>(r[i], i));
  for (int i = 1; i <= k; i++) {
    for (auto it : u[i]) s.insert(it);
    if (!s.size()) {
      puts("-1");
      return 0;
    }
    auto val = *s.begin();
    rk[val.second] = i;
    s.erase(s.begin());
    while ((val = *s.begin()).first == i) {
      rk[val.second] = i;
      s.erase(s.begin());
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", rk[i]);
  return 0;
}