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 pii = pair<int, int>;
const int N = (int)2e5 + 10;
int n, m, k;
int value[N];
int used[N];
int initial[N];
vector<int> g[N], gt[N];
vector<int> top_sort;
set<int> with_rank[N];
void fail() {
  puts("-1");
  exit(0);
}
bool has_cycle(int v) {
  used[v] = 1;
  for (int to : g[v]) {
    if (!used[to] && has_cycle(to))
      return true;
    else if (used[to] == 1)
      return true;
  }
  used[v] = 2;
  return false;
}
bool has_cycle() {
  fill(used, used + n, 0);
  for (int i = 0; i < n; i++) {
    if (!used[i] && has_cycle(i)) return true;
  }
  return false;
}
void build_top(int v) {
  used[v] = 1;
  for (int to : g[v]) {
    if (!used[to]) build_top(to);
  }
  top_sort.push_back(v);
}
void build_top() {
  fill(used, used + n, 0);
  for (int i = 0; i < n; i++) {
    if (!used[i]) build_top(i);
  }
  reverse(top_sort.begin(), top_sort.end());
}
void put_ranks() {
  for (int v : top_sort) {
    int max_value = k;
    for (int from : gt[v]) max_value = min(max_value, value[from] - 1);
    if (max_value <= 0 || (value[v] != 0 && value[v] > max_value)) fail();
    if (value[v] == 0) value[v] = max_value;
  }
}
bool cmp(int a, int b) {
  if (value[a] != value[b]) return value[a] < value[b];
  return a < b;
}
set<int, bool (*)(int, int)> keksi(cmp);
int out_deg[N];
void try_add_to_keksi(int v) {
  if (out_deg[v] > 0 || initial[v] != 0) return;
  keksi.insert(v);
}
void find_keks(int level) {
  auto best = keksi.begin();
  if (best == keksi.end()) return;
  int best_id = *best;
  keksi.erase(best);
  with_rank[value[best_id]].erase(best_id);
  value[best_id] = level;
  with_rank[level].insert(best_id);
}
void decrease_deg(int v) {
  out_deg[v]--;
  try_add_to_keksi(v);
}
void solve() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &value[i]);
    initial[i] = value[i];
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    g[a].push_back(b);
    gt[b].push_back(a);
    out_deg[a]++;
  }
  if (has_cycle()) fail();
  build_top();
  put_ranks();
  for (int i = 0; i < n; i++) {
    with_rank[value[i]].insert(i);
    try_add_to_keksi(i);
  }
  for (int i = 1; i <= k; i++) {
    if (with_rank[i].size() == 0) find_keks(i);
    for (auto v : with_rank[i]) {
      keksi.erase(v);
      for (auto from : gt[v]) decrease_deg(from);
    }
    if (with_rank[i].size() == 0) fail();
  }
  for (int i = 0; i < n; i++) printf("%d ", value[i]);
  puts("");
}
int main() {
  solve();
  (void)0;
  return 0;
}