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;
template <typename S, typename T>
void xmin(S& a, T const& b) {
  if (b < a) a = b;
}
template <typename S, typename T>
void xmax(S& a, T const& b) {
  if (b > a) a = b;
}
vector<vector<int>> g, rg;
vector<int> r, mir, mar;
vector<char> vis;
void rec(int u) {
  if (vis[u]) return;
  vis[u] = 1;
  for (int e : g[u]) {
    rec(e);
    xmax(mir[u], mir[e] + 1);
  }
}
void rec2(int u) {
  if (vis[u]) return;
  vis[u] = 1;
  for (int e : rg[u]) {
    rec2(e);
    xmin(mar[u], mar[e] - 1);
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  g.assign(n, vector<int>());
  rg.assign(n, vector<int>());
  r.assign(n, 0);
  mar.assign(n, k);
  mir.assign(n, 1);
  vector<int> rfree(k + 1, 1);
  rfree[0] = 0;
  for (int i = 0; i < n; ++i) {
    cin >> r[i];
    if (r[i] != 0) {
      mir[i] = mar[i] = r[i];
      rfree[r[i]] = 0;
    }
  }
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    rg[b].push_back(a);
  }
  vis.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    rec(i);
  }
  vis.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    rec2(i);
  }
  bool f = false;
  for (int i = 0; i < n; ++i) {
    if (mir[i] < 1 || mir[i] > k) f = true;
    if (mar[i] < 1 || mar[i] > k) f = true;
    if (mir[i] > mar[i]) f = true;
  }
  if (!f) {
    vector<vector<int>> bumi(k + 1), buma(k + 1);
    for (int i = 0; i < n; ++i) {
      if (r[i] == 0) bumi[mir[i]].push_back(i);
      if (r[i] == 0) buma[mar[i]].push_back(i);
    }
    priority_queue<pair<int, int>, vector<pair<int, int>>,
                   greater<pair<int, int>>>
        ac;
    for (int i = 1; i <= k; ++i) {
      for (int e : bumi[i]) {
        ac.emplace(mar[e], e);
      }
      for (auto const& e : buma[i]) {
        if (r[e] == 0) {
          r[e] = i;
          rfree[i] = 0;
        }
      }
      if (!rfree[i]) continue;
      while (!ac.empty() && r[ac.top().second] != 0) ac.pop();
      if (ac.empty()) {
        f = true;
        break;
      } else {
        int u = ac.top().second;
        ac.pop();
        r[u] = i;
      }
    }
  }
  if (!f) {
    for (int i = 0; i < n; ++i) {
      assert(r[i] <= mar[i]);
      assert(r[i] >= mir[i]);
      for (int e : g[i]) {
        if (r[e] >= r[i]) f = true;
      }
    }
  }
  fill(rfree.begin(), rfree.end(), 0);
  for (int i = 0; i < n; ++i) {
    rfree[r[i]] = 1;
  }
  if (!f) {
    assert(accumulate(rfree.begin(), rfree.end(), 0) == k);
  }
  if (f) {
    cout << "-1\n";
  } else {
    copy(r.begin(), r.end(), ostream_iterator<int>(cout, " "));
    cout << "\n";
  }
  return 0;
}