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>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const int N = 2e5 + 10;
int n, m, k, cnt, head[N], a[N], up[N], dn[N], ans[N], tu[N], tv[N], in[N];
vector<int> g[N];
struct E {
  int to, next;
} edge[N << 1];
struct node {
  int val, pos;
  bool operator<(const node &x) const { return x.val > val; }
};
inline void add(int u, int v) {
  edge[++cnt] = {v, head[u]};
  head[u] = cnt;
  in[v]++;
}
priority_queue<node> q;
inline void topo(int opt) {
  queue<int> p;
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (!in[i]) p.push(i), cnt++;
  while (!p.empty()) {
    int u = p.front();
    p.pop();
    if (up[u] < dn[u]) printf("-1"), exit(0);
    for (int i = head[u]; i; i = edge[i].next) {
      int v = edge[i].to;
      if (opt)
        chmax(dn[v], dn[u] + 1);
      else
        chmin(up[v], up[u] - 1);
      if (!--in[v]) p.push(v), cnt++;
    }
  }
  if (cnt < n) printf("-1"), exit(0);
}
int main() {
  n = read(), m = read(), k = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (!a[i])
      up[i] = k, dn[i] = 1;
    else
      up[i] = dn[i] = a[i];
  }
  for (int i = 1; i <= m; i++) {
    tu[i] = read(), tv[i] = read();
    add(tu[i], tv[i]);
  }
  topo(0);
  memset(head, 0, sizeof(head));
  cnt = 0;
  for (int i = 1; i <= m; i++) add(tv[i], tu[i]);
  topo(1);
  for (int i = 1; i <= n; i++) g[up[i]].push_back(i);
  for (int i = k; i >= 1; i--) {
    for (int j = 0; j < g[i].size(); j++) q.push((node){dn[g[i][j]], g[i][j]});
    if (q.empty()) return printf("-1"), 0;
    int x = q.top().pos;
    ans[x] = i;
    q.pop();
    while (!q.empty()) {
      int x = q.top().pos;
      if (dn[x] < i) break;
      ans[x] = i;
      q.pop();
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
  ;
}