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;
inline int in() {
  int x, y;
  y = scanf("%d", &x);
  return x;
}
const int N = 300003;
int n, m, k;
vector<int> g[N], gr[N];
int to[N], from[N], mark[N], res[N], din[N], rem;
int x[N], r[N], y[N];
void out() {
  cout << "-1\n";
  exit(0);
}
void dfs(int v) {
  mark[v] = 1;
  for (int u : g[v])
    if (mark[u] == 1)
      out();
    else {
      if (!mark[u]) dfs(u);
      from[v] = max(from[v], from[u]);
    }
  from[v]++;
  if (r[v] > 0) {
    if (from[v] > r[v]) out();
    from[v] = r[v];
  }
  if (from[v] > k) out();
  mark[v] = 2;
}
bool seen[N];
set<pair<int, int>> ss, qs;
void updateRem() {
  while (rem > 0 && seen[rem]) {
    rem--;
    while (qs.size() && qs.rbegin()->first >= rem) {
      int v = qs.rbegin()->second;
      qs.erase({to[v], v});
      ss.insert({from[v], v});
    }
  }
}
void addSource(int v) {
  int x = to[v];
  if (x >= rem)
    ss.insert({from[v], v});
  else
    qs.insert({x, v});
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) r[i] = in();
  for (int i = 0; i < m; i++) {
    x[i] = in() - 1;
    y[i] = in() - 1;
    g[x[i]].push_back(y[i]);
    gr[y[i]].push_back(x[i]);
    din[y[i]]++;
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) dfs(i);
  rem = k;
  fill(to, to + n, k);
  for (int i = 0; i < n; i++) {
    if (r[i] > 0) to[i] = min(to[i], r[i]);
    if (!din[i]) addSource(i);
  }
  for (int i = 0; i < n; i++) {
    if (ss.empty()) out();
    int v = ss.rbegin()->second;
    ss.erase({from[v], v});
    assert(to[v] >= rem);
    if (r[v] > 0)
      res[v] = r[v];
    else
      res[v] = max(rem, from[v]);
    seen[res[v]] = true;
    updateRem();
    for (int u : g[v]) {
      din[u]--;
      to[u] = min(to[u], res[v] - 1);
      if (to[u] < from[u]) out();
      if (din[u] == 0) addSource(u);
    }
  }
  if (rem != 0) out();
  for (int i = 0; i < m; i++)
    if (res[x[i]] <= res[y[i]]) out();
  for (int i = 0; i < n; i++) cout << res[i] << " ";
  cout << endl;
}