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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int N = 200 * 1000 + 10;
vector<int> g[N];
vector<int> gr[N];
int col[N];
int init_rank[N];
int ans_rank[N];
int L[N];
int R[N];
bool is_covered[N];
bool has_cyc;
void dfs_cyc(int v) {
  col[v] = 1;
  for (int to : g[v]) {
    if (col[to] == 0) {
      dfs_cyc(to);
      continue;
    }
    if (col[to] == 1) has_cyc = true;
  }
  col[v] = 2;
}
int calc_dp(int v) {
  if (col[v]) return L[v];
  col[v] = 1;
  for (int to : g[v]) L[v] = max(L[v], calc_dp(to) + 1);
  return L[v];
}
int calc_dp_rev(int v) {
  if (col[v]) return R[v];
  col[v] = 1;
  for (int to : gr[v]) R[v] = min(R[v], calc_dp_rev(to) - 1);
  return R[v];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    cin >> init_rank[i];
    is_covered[init_rank[i]] = true;
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    g[x].push_back(y);
    gr[y].push_back(x);
  }
  fill(L, L + n, 1);
  fill(R, R + n, k);
  for (int i = 0; i < n; i++)
    if (init_rank[i]) L[i] = R[i] = init_rank[i];
  for (int i = 0; i < n; i++)
    if (col[i] == 0) dfs_cyc(i);
  if (has_cyc) {
    cout << -1 << endl;
    return 0;
  }
  fill(col, col + n, 0);
  for (int i = 0; i < n; i++) calc_dp(i);
  fill(col, col + n, 0);
  for (int i = 0; i < n; i++) calc_dp_rev(i);
  bool ok = true;
  for (int i = 0; i < n; i++) {
    if (L[i] > R[i]) ok = false;
  }
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) ans_rank[i] = init_rank[i] ? init_rank[i] : R[i];
  set<pii> open;
  vector<vector<pii>> open_at(k + 1);
  vector<vector<pii>> close_at(k + 1);
  for (int i = 0; i < n; i++) {
    open_at[L[i]].emplace_back(R[i], i);
    close_at[R[i]].emplace_back(R[i], i);
  }
  for (int i = 1; i <= k; i++) {
    for (auto p : open_at[i]) open.insert(p);
    if (!is_covered[i]) {
      if (open.empty()) {
        cout << -1 << endl;
        return 0;
      }
      auto it = open.begin();
      ans_rank[it->second] = i;
      open.erase(it);
    }
    for (auto p : close_at[i]) open.erase(p);
  }
  for (int i = 0; i < n; i++) cout << ans_rank[i] << " ";
  cout << endl;
}