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>
const int MAXN = 300005;
using namespace std;
int n, m, k, R[MAXN], hi[MAXN], lo[MAXN], deg[MAXN], ans[MAXN];
vector<int> Gr[MAXN], G[MAXN];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> R[i];
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    Gr[v].push_back(u);
    G[u].push_back(v);
    deg[v]++;
  }
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 0) {
      Q.push(i);
    }
  }
  int vis = 0;
  vector<int> ord;
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    vis++;
    ord.push_back(v);
    for (int to : G[v]) {
      deg[to]--;
      if (deg[to] == 0) {
        Q.push(to);
      }
    }
  }
  if (vis != n) {
    cout << "-1\n";
    return;
  }
  for (int i = n - 1; i >= 0; i--) {
    int v = ord[i];
    lo[v] = (R[v] == 0 ? 1 : R[v]);
    for (int to : G[v]) {
      lo[v] = max(lo[v], lo[to] + 1);
    }
  }
  for (int i = 0; i < n; i++) {
    int v = ord[i];
    hi[v] = (R[v] == 0 ? k : R[v]);
    for (int fr : Gr[v]) {
      hi[v] = min(hi[v], hi[fr] - 1);
    }
  }
  int ok = 1;
  vector<pair<int, int> > add;
  for (int i = 1; i <= n; i++) {
    if (hi[i] < lo[i] || lo[i] > k || hi[i] < 1) {
      ok = 0;
      break;
    }
    add.push_back(pair<int, int>(hi[i], i));
  }
  sort((add).begin(), (add).end());
  if (!ok) {
    cout << "-1\n";
    return;
  }
  set<pair<int, int> > P;
  for (int i = k; i >= 1; i--) {
    while (!add.empty() && add.back().first >= i) {
      int v = add.back().second;
      P.insert(pair<int, int>(lo[v], v));
      add.pop_back();
    }
    if (P.empty()) {
      ok = 0;
      break;
    }
    int cnt = 0;
    while (!P.empty() && (P.rbegin()->first == i || cnt == 0)) {
      pair<int, int> p = *P.rbegin();
      ans[p.second] = i;
      cnt++;
      P.erase(--P.end());
    }
  }
  if (!ok) {
    cout << "-1\n";
    return;
  }
  for (int i = 1; i <= n; i++) {
    for (int to : G[i]) {
      assert(ans[i] >= ans[to]);
    }
    if (R[i] != 0) {
      assert(ans[i] == R[i]);
    }
    cout << ans[i] << " \n"[i == n];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  for (int i = 1; i <= tc; i++) {
    solve();
  }
  return 0;
}