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 namespace placeholders;
template <class T>
void mini(T &l, T r) {
  l = min(l, r);
}
template <class T>
void maxi(T &l, T r) {
  l = max(l, r);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const int M = 2E5 + 10, N = M;
int n, m, K;
int rk[M], lb[M], ub[M], deg[M];
vector<int> E[M], b, c[N];
int h[M];
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", rk + i);
    if (rk[i])
      lb[i] = ub[i] = rk[i];
    else
      lb[i] = 1, ub[i] = K;
  }
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    ++deg[v];
  }
  queue<int> Q;
  for (int i = 1; i <= n; ++i)
    if (deg[i] == 0) Q.push(i);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    b.push_back(u);
    for (auto &v : E[u])
      if (--deg[v] == 0) Q.push(v);
  }
  if (b.size() < n) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    int u = b[i];
    for (auto &v : E[u]) ub[v] = min(ub[v], ub[u] - 1);
  }
  for (int i = n - 1; i >= 0; --i) {
    int u = b[i];
    for (auto &v : E[u]) lb[u] = max(lb[u], lb[v] + 1);
  }
  for (int i = 1; i <= n; ++i) {
    if (lb[i] > ub[i]) {
      puts("-1");
      return 0;
    }
    c[ub[i]].push_back(i);
  }
  priority_queue<pair<int, int> > Q1;
  for (int i = K; i >= 1; --i) {
    for (auto &t : c[i]) Q1.emplace(lb[t], t);
    if (Q1.empty()) {
      puts("-1");
      return 0;
    }
    int u = Q1.top().second;
    rk[u] = i;
    Q1.pop();
    for (; !Q1.empty() && Q1.top().first == i; Q1.pop())
      rk[Q1.top().second] = i;
  }
  for (int i = 1; i <= n; ++i) printf("%d%c", rk[i], "\n "[i < n]);
  return 0;
}