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 db = double;
using vi = vector<int>;
const int maxn = 200006;
int n, m, k;
int r[maxn];
vi e[maxn];
vi g[maxn];
int q[maxn];
int d[maxn];
int tot;
int oud[maxn];
int ans[maxn];
vi sur[maxn];
bool v[maxn];
set<pair<int, int>> S;
void dfs(int x) {
  v[x] = true;
  for (auto i : e[x])
    if (!v[i]) dfs(i);
  q[++tot] = x;
}
void Del(int x) {
  for (auto j : g[x]) {
    oud[j]--;
    if (oud[j] == 0 && r[j] == 0) S.insert({d[j], j});
  }
  S.erase(make_pair(d[x], x));
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &r[i]);
    sur[r[i]].push_back(i);
  }
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[y].push_back(x);
    g[x].push_back(y);
  }
  memset(v, false, sizeof(v));
  tot = 0;
  for (int i = 1; i <= n; ++i)
    if (!v[i]) dfs(i);
  for (int i = 1; i <= n; ++i) d[i] = 1;
  reverse(q + 1, q + n + 1);
  for (int u = 1; u <= n; ++u) {
    int i = q[u];
    d[i] = max(d[i], r[i]);
    for (auto j : e[i]) d[j] = max(d[j], d[i] + 1);
  }
  for (int i = 1; i <= n; ++i) oud[i] = e[i].size();
  for (int i = 1; i <= n; ++i)
    if (oud[i] == 0 && r[i] == 0) S.insert({d[i], i});
  for (int i = k; i >= 1; --i) {
    bool flag = false;
    for (auto x : sur[i]) {
      flag = true;
      ans[x] = i;
      Del(x);
    }
    while (!S.empty() && S.rbegin()->first >= i) {
      flag = true;
      int x = S.rbegin()->second;
      ans[x] = i;
      Del(x);
    }
    if (!flag) {
      if (S.empty()) {
        puts("-1");
        return 0;
      }
      int x = S.rbegin()->second;
      ans[x] = i;
      Del(x);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!(ans[i] >= 1 && ans[i] <= k && (r[i] == 0 || ans[i] == r[i]))) {
      puts("-1");
      return 0;
    }
  for (int i = 1; i <= n; ++i)
    for (auto j : e[i])
      if (ans[i] >= ans[j]) {
        puts("-1");
        return 0;
      }
  set<int> ss;
  for (int i = 1; i <= n; ++i) ss.insert(ans[i]);
  if ((int)ss.size() != k) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  puts("");
}