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;
template <typename T>
inline bool cmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool cmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int x = 0, f = 1;
  char ch;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return x * f;
}
const int MaxN = 201234;
int n, m, K, on, l[MaxN], r[MaxN], deg[MaxN], ord[MaxN], ans[MaxN];
vector<int> e[MaxN], val[MaxN];
queue<int> q;
int main() {
  int i;
  n = read();
  m = read();
  K = read();
  for (i = 1; i <= n; i++) {
    r[i] = read();
    if (!r[i])
      l[i] = 1, r[i] = K;
    else
      l[i] = r[i];
  }
  for (i = 1; i <= m; i++) {
    int u = read(), v = read();
    e[u].emplace_back(v);
    ++deg[v];
  }
  for (i = 1; i <= n; i++)
    if (!deg[i]) q.push(i);
  while (!q.empty()) {
    int c = q.front();
    q.pop();
    ord[++on] = c;
    for (auto y : e[c])
      if (!--deg[y]) q.push(y);
  }
  if (on != n) return puts("-1"), 0;
  for (i = 1; i <= n; i++) {
    int x = ord[i];
    for (auto y : e[x]) cmin(r[y], r[x] - 1);
  }
  for (i = n; i; i--) {
    int x = ord[i];
    for (auto y : e[x]) cmax(l[x], l[y] + 1);
  }
  for (i = 1; i <= n; i++)
    if (l[i] > r[i]) return puts("-1"), 0;
  for (i = 1; i <= n; i++) val[l[i]].push_back(i);
  set<pair<int, int> > s;
  for (i = 1; i <= K; i++) {
    for (auto x : val[i]) s.insert(make_pair(r[x], x));
    if (s.empty()) return puts("-1"), 0;
    pair<int, int> v = *s.begin();
    for (ans[v.second] = i, s.erase(s.begin());
         !s.empty() && (v = *s.begin()).first == i;
         ans[v.second] = i, s.erase(s.begin()))
      ;
  }
  for (i = 1; i <= n; i++)
    for (auto y : e[i])
      if (ans[y] >= ans[i]) return puts("-1"), 0;
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}