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 INF = 0x3f3f3f3f;
const int N = 2e5 + 5;
int n, m, k;
vector<int> e[N], add[N];
int d[N];
int ans[N], up[N], down[N];
int ql, qr, q[N];
set<pair<int, int> > st;
bool solve() {
  if (n < k) return false;
  ql = qr = 0;
  for (int i = 1; i <= n; ++i)
    if (d[i] == 0) q[++qr] = i;
  while (ql ^ qr) {
    int u = q[++ql];
    int len = e[u].size();
    for (int i = 0; i < len; ++i) {
      int v = e[u][i];
      --d[v];
      if (!d[v]) q[++qr] = v;
    }
  }
  if (qr ^ n) return false;
  for (int i = 1; i <= n; ++i) {
    int u = q[i];
    for (auto v : e[u]) up[v] = min(up[v], up[u] - 1);
  }
  for (int i = n; i >= 1; --i) {
    int u = q[i];
    for (auto v : e[u]) down[u] = max(down[u], down[v] + 1);
    if (down[u] > up[u]) return false;
  }
  for (int i = 1; i <= n; ++i) add[down[i]].push_back(i);
  for (int i = 1; i <= k; ++i) {
    for (auto v : add[i]) st.insert(make_pair(up[v], v));
    if (st.empty()) return false;
    int u = (*st.begin()).second;
    st.erase(st.begin());
    ans[u] = i;
    while (!st.empty() && (*st.begin()).first == i)
      ans[(*st.begin()).second] = i, st.erase(st.begin());
  }
  return true;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  st.clear();
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 0)
      down[i] = 1, up[i] = k;
    else
      down[i] = up[i] = x;
    e[i].clear();
    d[i] = 0;
  }
  for (int i = 1; i <= k; ++i) add[i].clear();
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    ++d[v];
  }
  if (solve()) {
    for (int i = 1; i <= n; ++i) printf("%d%s", ans[i], i == n ? "\n" : " ");
  } else
    printf("-1\n");
  return 0;
}