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;
struct Graph {
  vector<vector<int> > adj;
  vector<vector<int> > radj;
  vector<int> FV;
  vector<int> UP, LO;
  Graph(int N) : adj(N), radj(N), FV(N, -1), UP(N, -1), LO(N, -1) {}
  void addEdge(int u, int v) {
    adj[u].push_back(v);
    radj[v].push_back(u);
  }
  bool build(int K) {
    vector<int> VIS(adj.size());
    for (int i = 0; i < adj.size(); ++i) {
      if (!VIS[i] && hasCycle(i, VIS)) return false;
    }
    for (int i = 0; i < adj.size(); ++i)
      if (UP[i] == -1 && !buildUpper(i, K)) return false;
    for (int i = 0; i < adj.size(); ++i)
      if (LO[i] == -1 && !buildLower(i, K)) return false;
    return true;
  }
  bool hasCycle(int u, vector<int> &VIS) {
    if (VIS[u] == 2) return false;
    if (VIS[u] == 1) return true;
    VIS[u] = 1;
    for (int v : adj[u]) {
      if (hasCycle(v, VIS)) return true;
    }
    VIS[u] = 2;
    return false;
  }
  bool buildUpper(int u, int K) {
    int lowest = FV[u] == -1 ? K : FV[u];
    for (int v : adj[u]) {
      if (UP[v] == -1) {
        if (!buildUpper(v, K)) return false;
      }
      lowest = min(UP[v] - 1, lowest);
    }
    if (FV[u] != -1 && FV[u] > lowest) {
      return false;
    }
    UP[u] = lowest;
    return true;
  }
  bool buildLower(int u, int K) {
    int biggest = FV[u] == -1 ? 1 : FV[u];
    for (int v : radj[u]) {
      if (LO[v] == -1) {
        if (!buildLower(v, K)) return false;
      }
      biggest = max(biggest, LO[v] + 1);
    }
    if (FV[u] != -1 && FV[u] < biggest) {
      return false;
    }
    LO[u] = biggest;
    return true;
  }
};
struct Range {
  int s, e, id;
};
bool byEnd(Range a, Range b) { return a.e < b.e; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, M, K;
  cin >> N >> M >> K;
  Graph G(N);
  for (int i = 0; i < N; ++i) {
    int n;
    cin >> n;
    if (n == 0) continue;
    G.FV[i] = K - n + 1;
  }
  for (int i = 0; i < M; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    G.addEdge(u, v);
  }
  if (!G.build(K)) {
    cout << -1 << '\n';
    return 0;
  }
  vector<Range> R;
  for (int i = 0; i < N; ++i) {
    R.push_back({G.LO[i], G.UP[i], i});
    if (G.LO[i] > G.UP[i]) {
      cout << -1 << '\n';
      return 0;
    }
  }
  set<int> S;
  for (int i = 1; i <= K; ++i) {
    S.insert(i);
  }
  sort(R.begin(), R.end(), byEnd);
  vector<int> ANS(N);
  for (int i = 0; i < R.size(); ++i) {
    auto l = S.lower_bound(R[i].s);
    if (l == S.end() || *l > R[i].e) {
      ANS[R[i].id] = R[i].e;
    } else {
      ANS[R[i].id] = *l;
      S.erase(*l);
    }
  }
  if (S.size()) {
    cout << -1 << '\n';
    return 0;
  }
  for (int v : ANS) {
    cout << K - v + 1 << " ";
  }
  cout << '\n';
  return 0;
}