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 = 1 << 18;
int n, m, k;
vector<int> e[N], g[N];
int vis[N], tmr;
int r[N];
vector<int> vct;
void dfs(int v) {
  vis[v] = tmr;
  for (int to : e[v]) {
    if (vis[to] != tmr) {
      dfs(to);
    }
  }
  vct.push_back(v);
}
int main() {
  while (scanf("%d%d%d", &n, &m, &k) == 3) {
    for (int i = 0; i < n; i++) {
      e[i].clear();
      g[i].clear();
    }
    for (int i = 0; i < n; i++) scanf("%d", &r[i]);
    for (int i = 0; i < m; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      --u;
      --v;
      e[u].push_back(v);
      g[v].push_back(u);
    }
    vct.clear();
    tmr++;
    for (int i = 0; i < n; i++) {
      if (vis[i] != tmr) {
        dfs(i);
      }
    }
    reverse(vct.begin(), vct.end());
    tmr++;
    bool fail = 0;
    vector<int> d(n);
    vector<int> rr(n);
    for (int i = 0; i < n; i++) rr[i] = r[i];
    for (int i = n - 1; i >= 0; i--) {
      int v = vct[i];
      int mx = 0;
      vis[v] = tmr;
      for (int to : e[v]) {
        if (vis[to] != tmr) fail = 1;
        mx = max(mx, r[to]);
        d[v] = max(d[v], d[to] + 1);
      }
      if (r[v] == 0) {
        r[v] = mx + 1;
        fail |= r[v] > k;
      } else {
        fail |= mx >= r[v];
      }
    }
    if (!fail) {
      vector<pair<int, int> > b;
      for (int i = 0; i < n; i++) b.push_back({r[i], i});
      sort(b.begin(), b.end());
      set<int> st;
      for (int i = 1; i <= k; i++) st.insert(i);
      for (int i = 0; i < n; i++) st.erase(rr[i]);
      reverse(b.begin(), b.end());
      for (auto o : b) {
        int v = o.second;
        if (rr[v] != 0) continue;
        int mn = k + 1;
        for (int to : g[v]) {
          mn = min(mn, r[to]);
        }
        auto it = st.lower_bound(mn);
        if (it != st.begin() && *--it >= r[v]) {
          r[v] = *it;
          st.erase(it);
        }
      }
      fail |= !st.empty();
    }
    if (fail)
      cout << -1 << endl;
    else
      for (int i = 0; i < n; i++) cout << r[i] << " \n"[i + 1 == n];
  }
  return 0;
}