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, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
const int MAXN = 220000;
vector<int> adj[MAXN];
int deg[MAXN], a[MAXN], b[MAXN];
int id[MAXN], ans[MAXN];
void add_edge(int u, int v) {
  adj[u].push_back(v);
  deg[v]++;
}
int q[MAXN];
bool cmp(int u, int v) {
  if (a[u] != a[v]) return a[u] < a[v];
  return b[u] < b[v];
}
set<pair<int, int> > S;
int main() {
  int n, m, K;
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    if (a[i])
      b[i] = a[i];
    else
      b[i] = K;
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    if (a[u] && a[v] && a[u] <= a[v]) return puts("-1"), 0;
  }
  for (int i = 1; i <= n; i++)
    if (!a[i]) a[i] = 1, b[i] = K;
  int qn = 0;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q[qn++] = i;
  for (int it = 0; it < qn; it++) {
    int u = q[it];
    for (int j = 0; j < ((int)(adj[u].size())); j++) {
      int v = adj[u][j];
      if (--deg[v] == 0) q[qn++] = v;
    }
  }
  if (qn != n) {
    return puts("-1"), 0;
  }
  for (int i = 0; i < n; i++) {
    int u = q[i];
    for (int j = 0; j < ((int)(adj[u].size())); j++) {
      int v = adj[u][j];
      chkmin(b[v], b[u] - 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int u = q[i];
    for (int j = 0; j < ((int)(adj[u].size())); j++) {
      int v = adj[u][j];
      chkmax(a[u], a[v] + 1);
    }
  }
  for (int i = 1; i <= n; i++)
    if (a[i] > b[i]) return puts("-1"), 0;
  for (int i = 1; i <= n; i++) id[i] = i;
  sort(id + 1, id + n + 1, cmp);
  for (int i = 1, j = 1; i <= K; i++) {
    while (j <= n && a[id[j]] <= i)
      S.insert(pair<int, int>(b[id[j]], id[j])), j++;
    int flg = 0;
    while (!S.empty()) {
      set<pair<int, int> >::iterator it = S.begin();
      if (it->first > i) break;
      flg = 1;
      int u = it->second;
      ans[u] = i;
      S.erase(it);
    }
    if (!flg && S.empty()) return puts("-1"), 0;
    if (flg) continue;
    set<pair<int, int> >::iterator it = S.begin();
    int u = it->second;
    ans[u] = i;
    S.erase(it);
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}