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 long long inf = 1e9;
const long long INF = 1e18;
const long long MAX = 2e5 + 7;
long long n, m, k, num[MAX], col[MAX], remain[MAX];
long long mn[MAX], mx[MAX];
vector<long long> graph[MAX], st, rev[MAX];
bool pos;
void dfs(long long u) {
  if (col[u] == 1) {
    pos = false;
    return;
  }
  if (col[u] == 2) return;
  col[u] = 1;
  for (auto v : graph[u]) {
    dfs(v);
  }
  col[u] = 2;
  st.push_back(u);
}
long long ans[MAX];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  pos = true;
  for (long long i = 1; i <= n; i++) {
    cin >> num[i];
    remain[num[i]] = 1;
    ans[i] = num[i];
  }
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    rev[v].push_back(u);
  }
  for (long long i = 1; i <= n; i++) {
    if (!col[i]) dfs(i);
  }
  if (!pos) {
    cout << -1;
    return 0;
  }
  for (auto v : st) {
    long long cur = 1;
    for (auto u : graph[v]) cur = max(cur, mn[u] + 1);
    mn[v] = cur;
    if (num[v]) {
      if (mn[v] > num[v]) {
        cout << -1;
        return 0;
      }
      mn[v] = num[v];
    }
  }
  reverse(st.begin(), st.end());
  for (auto v : st) {
    long long cur = k;
    for (auto u : rev[v]) cur = min(cur, mx[u] - 1);
    mx[v] = cur;
    if (num[v]) {
      if (mx[v] < num[v]) {
        cout << -1;
        return 0;
      }
      mx[v] = num[v];
    }
  }
  vector<pair<long long, long long>> open[k + 1];
  for (long long i = 1; i <= n; i++) {
    if (mn[i] > mx[i] or mn[i] < 0 or mx[i] > k) {
      cout << -1;
      return 0;
    }
    ans[i] = mx[i];
    open[mn[i]].push_back({mx[i], i});
  }
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      pq;
  for (long long i = 1; i <= k; i++) {
    for (auto v : open[i]) pq.push(v);
    while (!pq.empty() and pq.top().first < i) pq.pop();
    if (remain[i]) continue;
    if (pq.empty()) {
      cout << -1;
      return 0;
    }
    auto cur = pq.top();
    pq.pop();
    ans[cur.second] = i;
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}