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>
const int MAXN = 2e5 + 5;
namespace Heap {
std::pair<int, int> pool[MAXN];
int tot;
void init() { tot = 0; }
std::pair<int, int> top() { return pool[1]; }
void push(int x, int y) {
  pool[++tot] = std::make_pair(x, y);
  std::push_heap(pool + 1, pool + tot + 1,
                 std::greater<std::pair<int, int> >());
}
void pop() {
  std::pop_heap(pool + 1, pool + tot + 1, std::greater<std::pair<int, int> >());
  --tot;
}
bool empty() { return tot == 0; }
}  // namespace Heap
int n, m, k;
int L[MAXN], R[MAXN];
std::vector<int> G[MAXN], vL[MAXN];
int ideg[MAXN];
int que[MAXN], hd = 0, tl = 0;
int ans[MAXN];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    static int x;
    scanf("%d", &x);
    if (x)
      L[i] = R[i] = x;
    else
      L[i] = 1, R[i] = k;
  }
  for (int i = 1; i <= m; ++i) {
    static int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    ++ideg[y];
  }
  for (int i = 1; i <= n; ++i)
    if (!ideg[i]) que[++tl] = i;
  while (hd != tl) {
    int now = que[++hd];
    for (int to : G[now])
      if (!--ideg[to]) que[++tl] = to;
  }
  if (tl != n) return puts("-1"), 0;
  for (int i = 1; i <= n; ++i) {
    int now = que[i];
    for (int to : G[now]) R[to] = std::min(R[to], R[now] - 1);
  }
  for (int i = n; i; --i) {
    int now = que[i];
    for (int to : G[now]) L[now] = std::max(L[now], L[to] + 1);
  }
  for (int i = 1; i <= n; ++i) {
    if (L[i] > R[i]) return puts("-1"), 0;
    vL[L[i]].push_back(i);
  }
  for (int i = 1; i <= k; ++i) {
    for (int x : vL[i]) Heap::push(R[x], x);
    if (Heap::empty()) return puts("-1"), 0;
    ans[Heap::top().second] = i;
    Heap::pop();
    while (!Heap::empty() && Heap::top().first == i) {
      ans[Heap::top().second] = i;
      Heap::pop();
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
  return 0;
}