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;
struct node {
  int v, next;
} t[Maxn];
struct N {
  int x, y;
  N(int a, int b) { x = a, y = b; }
};
int l[Maxn], r[Maxn], degre[Maxn], q[Maxn], head[Maxn], ans[Maxn];
int cnt;
vector<int> add[Maxn];
set<pair<int, int> > s;
void Insert(int u, int v) {
  cnt++;
  degre[v]++;
  t[cnt].v = v;
  t[cnt].next = head[u];
  head[u] = cnt;
}
int main() {
  int n, m, k, x, y, ql = 0, qr = 0;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (!x)
      l[i] = 1, r[i] = k;
    else
      l[i] = r[i] = x;
  }
  for (int i = 1; i <= m; i++) scanf("%d%d", &x, &y), Insert(x, y);
  for (int i = 1; i <= n; i++)
    if (!degre[i]) q[++qr] = i;
  while (ql != qr) {
    int x = q[++ql];
    for (int i = head[x]; i; i = t[i].next) {
      degre[t[i].v]--;
      if (!degre[t[i].v]) q[++qr] = t[i].v;
    }
  }
  if (qr != n) return 0 * puts("-1");
  for (int i = 1; i <= n; i++) {
    int x = q[i];
    for (int j = head[x]; j; j = t[j].next)
      r[t[j].v] = min(r[t[j].v], r[x] - 1);
  }
  for (int i = n; i >= 1; i--) {
    int x = q[i];
    for (int j = head[x]; j; j = t[j].next) l[x] = max(l[x], l[t[j].v] + 1);
  }
  for (int i = 1; i <= n; i++)
    if (l[i] > r[i]) return 0 * puts("-1");
  for (int i = 1; i <= n; i++) add[l[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < add[i].size(); j++)
      s.insert(make_pair(r[add[i][j]], add[i][j]));
    if (s.empty()) return 0 * puts("-1");
    set<pair<int, int> >::iterator t = s.begin();
    ans[t->second] = i;
    s.erase(s.begin());
    for (; !s.empty() && ((t = s.begin())->first) == i;)
      ans[t->second] = i, s.erase(s.begin());
  }
  for (int u = 1; u <= n; u++)
    for (int v = head[u]; v; v = t[v].next)
      if (ans[u] <= ans[t[v].v]) return 0 * puts("-1");
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], (i == n) ? '\n' : ' ');
  return 0;
}