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;
const int mv = 2e5 + 20, me = mv;
int adj[mv], to[me], ant[me], d[mv], z;
int d2[mv];
vector<int> par[mv];
int memo[mv];
set<int> st;
int r[mv];
inline void add(int a, int b) {
  to[z] = b, ant[z] = adj[a], adj[a] = z++;
  d[b]++;
}
bool ok(int n) {
  queue<int> q;
  for (int i = 0, __ = n; i < __; i++) {
    if (d[i] == 0) {
      q.push(i);
    }
  }
  memcpy(d2, d, sizeof d);
  int cnt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    cnt++;
    for (int i = adj[u]; i > -1; i = ant[i]) {
      d[to[i]]--;
      if (d[to[i]] == 0) {
        q.push(to[i]);
      }
    }
  }
  memcpy(d, d2, sizeof d);
  return cnt == n;
}
int length(int u) {
  if (memo[u] != -1) return memo[u];
  int ans = 1;
  ans = max(ans, r[u]);
  for (int i = adj[u]; i > -1; i = ant[i]) {
    ans = max(ans, length(to[i]) + 1);
  }
  return memo[u] = ans;
}
struct cmp {
  bool operator()(int a, int b) const { return length(a) < length(b); }
};
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  memset(memo, -1, sizeof memo);
  memset(adj, -1, sizeof adj);
  for (int i = 1, __ = k + 1; i < __; i++) {
    st.insert(i);
  }
  for (int i = 0, __ = n; i < __; i++) {
    cin >> r[i];
    st.erase(r[i]);
  }
  for (int i = 0, __ = m; i < __; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    par[v].push_back(u);
    add(u, v);
  }
  if (!ok(n)) {
    puts("-1");
    return 0;
  }
  priority_queue<int, vector<int>, cmp> fila;
  for (int u = 0, __ = n; u < __; u++) {
    if (par[u].empty()) {
      fila.push(u);
    }
  }
  bool ok = 1;
  while (!fila.empty()) {
    int u = fila.top();
    fila.pop();
    int mi = length(u);
    int ma = k;
    for (int p : par[u]) {
      ma = min(r[p] - 1, ma);
    }
    if (mi > ma) {
      ok = 0;
    }
    auto it = st.upper_bound(ma);
    if (r[u] == 0) {
      if (it == st.begin()) {
        r[u] = ma;
      } else {
        --it;
        if (*it < mi) {
          r[u] = ma;
        } else {
          r[u] = *it;
          st.erase(it);
        }
      }
    } else {
      if (r[u] < mi || r[u] > ma) {
        ok = 0;
      }
    }
    for (int i = adj[u]; i > -1; i = ant[i]) {
      d[to[i]]--;
      if (d[to[i]] == 0) {
        fila.push(to[i]);
      }
    }
  }
  if (st.size() > 0 || !ok) {
    puts("-1");
    return 0;
  }
  for (int i = 0, __ = n; i < __; i++) {
    if (i) cout << " ";
    cout << r[i];
  }
  cout << '\n';
}