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 N = 1e6 + 6;
vector<int> g[N], h[N];
int din[N], dout[N];
int n, m, k, r[N];
int ma[N], mi[N];
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", r + i);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    h[v].push_back(u);
    dout[u]++;
    din[v]++;
  }
  stack<int> st;
  for (int i = 1; i <= n; i++) {
    if (din[i] == 0) st.push(i);
  }
  vector<int> topo;
  while (!st.empty()) {
    int v = st.top();
    st.pop();
    topo.push_back(v);
    for (int u : g[v]) {
      din[u]--;
      if (din[u] == 0) st.push(u);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (r[i] == 0)
      ma[i] = k, mi[i] = 1;
    else
      ma[i] = mi[i] = r[i];
  }
  if (topo.size() != n) {
    puts("-1");
    return 0;
  }
  for (int v : topo) {
    for (int u : g[v]) {
      ma[u] = min(ma[u], ma[v] - 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int v = topo[i];
    for (int u : g[v]) mi[v] = max(mi[v], mi[u] + 1);
  }
  for (int i = 1; i <= n; i++) {
    if (ma[i] < mi[i]) {
      puts("-1");
      return 0;
    }
  }
  set<int> col;
  for (int i = 1; i <= k; i++) {
    col.insert(i);
  }
  set<pair<int, int> > sst;
  for (int i = 1; i <= n; i++)
    if (dout[i] == 0) {
      sst.insert(make_pair(ma[i], i));
    }
  while (!sst.empty()) {
    auto it = sst.begin();
    int v = it->second;
    sst.erase(it);
    r[v] = mi[v];
    for (int u : g[v]) {
      r[v] = max(r[v], r[u] + 1);
    }
    auto rm = col.lower_bound(r[v]);
    if (rm != col.end() && *rm <= ma[v]) {
      r[v] = *rm;
    }
    col.erase(r[v]);
    for (int u : h[v]) {
      dout[u]--;
      if (dout[u] == 0) sst.insert(make_pair(ma[u], u));
    }
  }
  if (!col.empty())
    puts("-1");
  else {
    for (int i = 1; i <= n; i++) printf("%d%c", r[i], i == n ? '\n' : ' ');
  }
  return 0;
}