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 std::make_pair;
using std::max;
using std::min;
using std::pair;
using std::queue;
using std::vector;
const int N = 200005;
int to[N], Next[N], last[N], l[N], r[N], edge, a[N], n, m, k, d[N], ans[N];
struct note {
  int x, y;
} e[N];
queue<int> q;
vector<int> v[N];
std::priority_queue<pair<int, int> > Q;
void add(int x, int y) {
  to[++edge] = y;
  Next[edge] = last[x];
  last[x] = edge;
}
void ter() {
  puts("-1");
  exit(0);
}
void solvemax() {
  memcpy(r, a, sizeof(a));
  for (int i = 1; i <= n; i++)
    if (!r[i]) r[i] = k;
  for (int i = 1; i <= m; i++) {
    add(e[i].x, e[i].y);
    d[e[i].y]++;
  }
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  int cnt = 0;
  while (!q.empty()) {
    cnt++;
    int x = q.front();
    q.pop();
    for (int i = last[x]; i; i = Next[i]) {
      int u = to[i];
      r[u] = min(r[u], r[x] - 1);
      d[u]--;
      if (!d[u]) q.push(u);
    }
  }
  if (cnt != n) ter();
}
void solvemin() {
  memcpy(l, a, sizeof(a));
  memset(last, 0, sizeof(last));
  edge = 0;
  for (int i = 1; i <= n; i++)
    if (!l[i]) l[i] = 1;
  for (int i = 1; i <= m; i++) {
    add(e[i].y, e[i].x);
    d[e[i].x]++;
  }
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = last[x]; i; i = Next[i]) {
      int u = to[i];
      l[u] = max(l[u], l[x] + 1);
      d[u]--;
      if (!d[u]) q.push(u);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%d%d", &e[i].x, &e[i].y);
  solvemax();
  solvemin();
  for (int i = 1; i <= n; i++)
    if ((r[i] < l[i]) || (a[i] && (a[i] != l[i] || a[i] != r[i]))) ter();
  for (int i = 1; i <= n; i++) v[r[i]].push_back(i);
  for (int i = k; i >= 1; i--) {
    for (auto j : v[i]) Q.push(make_pair(l[j], j));
    if (Q.empty()) ter();
    int x = Q.top().second;
    Q.pop();
    ans[x] = i;
    while (Q.size()) {
      int x = Q.top().second;
      if (l[x] < i) break;
      ans[x] = i;
      Q.pop();
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}