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 N = 200005;
struct P {
  int x, y, id;
  bool operator<(const P b) const { return y > b.y; }
} a[N];
bool cmp(P a, P b) { return a.x < b.x; }
vector<int> link[N];
bool vis[N];
int d[N];
int L[N], R[N];
int ans[N];
void dfs(int u) {
  vis[u] = 1;
  for (int v : link[u]) {
    if (!vis[v]) dfs(v);
    L[u] = max(L[u], L[v] + 1);
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x)
      L[i] = R[i] = x;
    else
      L[i] = 1, R[i] = k;
  }
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    link[x].push_back(y);
    d[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 0) dfs(i);
  queue<int> Q;
  for (int i = 1; i <= n; i++)
    if (d[i] == 0) Q.push(i);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int v : link[u]) {
      R[v] = min(R[v], R[u] - 1);
      d[v]--;
      if (d[v] == 0) Q.push(v);
    }
  }
  for (int i = 1; i <= n; i++)
    if (d[i]) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= n; i++)
    if (L[i] > R[i]) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= n; i++) a[i] = {L[i], R[i], i};
  sort(a + 1, a + n + 1, cmp);
  priority_queue<P> PQ;
  int j = 1;
  for (int i = 1; i <= k; i++) {
    while (j <= n && a[j].x <= i) PQ.push(a[j]), j++;
    bool gt = 0;
    while (!PQ.empty()) {
      P tmp = PQ.top();
      PQ.pop();
      if (tmp.y < i) continue;
      ans[tmp.id] = i;
      gt = 1;
      break;
    }
    if (!gt) {
      printf("-1\n");
      return 0;
    }
    while (!PQ.empty()) {
      P tmp = PQ.top();
      if (tmp.y == i) {
        ans[tmp.id] = i;
        PQ.pop();
      } else
        break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", ans[i]);
}