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;
const int Maxn = 200005;
int n, m, k;
int r[Maxn];
vector<int> neigh[Maxn], rneigh[Maxn];
int in[Maxn], out[Maxn];
set<int> S;
int mn[Maxn], mx[Maxn];
vector<int> add[Maxn];
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= k; i++) S.insert(i);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    if (r[i] > 0) {
      S.erase(r[i]);
      mn[i] = r[i];
      mx[i] = r[i];
    } else
      mn[i] = 1, mx[i] = k;
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    neigh[a].push_back(b);
    rneigh[b].push_back(a);
    out[a]++;
    in[b]++;
  }
  vector<int> seq;
  for (int i = 1; i <= n; i++)
    if (out[i] == 0) seq.push_back(i);
  for (int i = 0; i < seq.size(); i++) {
    int v = seq[i];
    if (r[v] > 0 && mn[v] > r[v]) {
      printf("-1\n");
      return 0;
    }
    for (int j = 0; j < rneigh[v].size(); j++) {
      int u = rneigh[v][j];
      mn[u] = max(mn[u], mn[v] + 1);
      if (--out[u] == 0) seq.push_back(u);
    }
  }
  if (seq.size() < n) {
    printf("-1\n");
    return 0;
  }
  seq.clear();
  for (int i = 1; i <= n; i++)
    if (in[i] == 0) seq.push_back(i);
  for (int i = 0; i < seq.size(); i++) {
    int v = seq[i];
    if (r[v] > 0 && mx[v] < r[v]) {
      printf("-1\n");
      return 0;
    }
    if (mx[v] < mn[v]) {
      printf("-1\n");
      return 0;
    }
    for (int j = 0; j < neigh[v].size(); j++) {
      int u = neigh[v][j];
      mx[u] = min(mx[u], mx[v] - 1);
      if (--in[u] == 0) seq.push_back(u);
    }
  }
  if (seq.size() < n) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) add[mx[i]].push_back(i);
  priority_queue<pair<int, int> > Q;
  for (int i = k; i > 0; i--) {
    for (int j = 0; j < add[i].size(); j++) {
      int ind = add[i][j];
      Q.push(pair<int, int>(mn[ind], ind));
    }
    if (S.find(i) != S.end()) {
      while (!Q.empty() && Q.top().first > i) {
        pair<int, int> v = Q.top();
        Q.pop();
        r[v.second] = v.first;
      }
      if (Q.empty()) {
        printf("-1\n");
        return 0;
      }
      pair<int, int> v = Q.top();
      Q.pop();
      r[v.second] = i;
    }
  }
  while (!Q.empty()) {
    pair<int, int> v = Q.top();
    Q.pop();
    r[v.second] = v.first;
  }
  for (int i = 1; i <= n; i++) printf("%d%c", r[i], i + 1 <= n ? ' ' : '\n');
  return 0;
}