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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 400005;
int n, m, k;
int a[N];
vector<int> g[N], gr[N];
int c[N];
bool dfs(int x) {
  c[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (c[h] == 1) return true;
    if (c[h] == 0) {
      if (dfs(h)) return true;
    }
  }
  c[x] = 2;
  return false;
}
int l[N];
void dfs1(int x) {
  c[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (!c[h]) dfs1(h);
    l[x] = max(l[x], l[h] + 1);
  }
}
int r[N];
void dfs2(int x) {
  c[x] = 1;
  for (int i = 0; i < gr[x].size(); ++i) {
    int h = gr[x][i];
    if (!c[h]) dfs2(h);
    r[x] = min(r[x], r[h] - 1);
  }
}
vector<int> v[N];
int ans[N];
void solv() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    gr[y].push_back(x);
  }
  for (int x = 1; x <= n; ++x) {
    if (c[x] == 0) {
      if (dfs(x)) {
        cout << "-1\n";
        return;
      }
    }
  }
  for (int x = 1; x <= n; ++x) {
    l[x] = 1;
    r[x] = k;
    if (a[x]) l[x] = r[x] = a[x];
  }
  memset(c, 0, sizeof c);
  for (int x = 1; x <= n; ++x) {
    if (!c[x]) {
      dfs1(x);
    }
  }
  memset(c, 0, sizeof c);
  for (int x = 1; x <= n; ++x) {
    if (!c[x]) {
      dfs2(x);
    }
  }
  for (int x = 1; x <= n; ++x) {
    if (l[x] > r[x]) {
      cout << "-1\n";
      return;
    }
  }
  for (int x = 1; x <= n; ++x) {
    v[r[x]].push_back(x);
  }
  set<pair<int, int> > s;
  for (int i = k; i >= 1; --i) {
    for (int j = 0; j < v[i].size(); ++j) {
      s.insert(make_pair(l[v[i][j]], v[i][j]));
    }
    while (1) {
      if (s.empty()) {
        cout << "-1\n";
        return;
      }
      int x = (--s.end())->second;
      s.erase((--s.end()));
      if (l[x] > i) {
        ans[x] = l[x];
        continue;
      }
      ans[x] = i;
      break;
    }
  }
  while (!s.empty()) {
    int x = (--s.end())->second;
    s.erase((--s.end()));
    ans[x] = l[x];
  }
  for (int x = 1; x <= n; ++x) cout << ans[x] << ' ';
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) {
    solv();
  }
  return 0;
}