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 INF = 0x3f3f3f3f;
const long long int llINF = (long long)(1e18) + 100;
const int MAXN = 2e5 + 10;
int n, m, k;
int minr[MAXN], deg[MAXN], color[MAXN], maxr[MAXN], vis[MAXN];
vector<int> adj[MAXN], order[MAXN];
bool ok;
void dfs(int v) {
  int mn = 1;
  vis[v] = 1;
  for (int x : adj[v]) {
    if (!vis[x]) dfs(x);
    mn = max(mn, minr[x] + 1);
  }
  if (minr[v] != 0) {
    if (mn > minr[v]) ok = 0;
  } else
    minr[v] = mn;
  if (mn > k) ok = 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> color[i];
    minr[i] = color[i];
  }
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    deg[y]++;
    adj[x].push_back(y);
  }
  memset(maxr, INF, sizeof maxr);
  for (int i = 1; i <= n; i++)
    if (minr[i] > 0) maxr[i] = minr[i];
  queue<int> fila;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) {
      fila.push(i);
      maxr[i] = min(maxr[i], k);
      order[maxr[i]].push_back(i);
    }
  int cnt = 0, vtx;
  while (!fila.empty()) {
    vtx = fila.front();
    fila.pop();
    cnt++;
    for (int x : adj[vtx]) {
      deg[x]--;
      maxr[x] = min(maxr[x], maxr[vtx] - 1);
      if (deg[x] == 0) {
        fila.push(x);
        if (maxr[x] <= 0) {
          cout << -1 << '\n';
          return 0;
        }
        order[maxr[x]].push_back(x);
      }
    }
  }
  if (cnt != n) {
    cout << -1 << '\n';
    return 0;
  }
  ok = 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  if (!ok) {
    cout << -1 << '\n';
    return 0;
  }
  priority_queue<pair<int, int> > pq;
  for (int rnk = k; rnk > 0; rnk--) {
    for (auto x : order[rnk]) pq.push({minr[x], x});
    if (pq.empty()) {
      cout << -1 << '\n';
      return 0;
    }
    do {
      color[pq.top().second] = rnk;
      pq.pop();
    } while (!pq.empty() and pq.top().first == rnk);
  }
  for (int i = 1; i <= n; i++) cout << color[i] << ' ';
  cout << '\n';
}