Manao is solving a problem with the following statement:

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He came up with a solution that produces the correct answers but is too slow. You are given the pseudocode of his solution, where the function getAnswer calculates the answer to the problem:
    
    
      
    getAnswer(a[1..n], b[1..len], h)  
      answer = 0  
      for i = 1 to n-len+1  
        answer = answer + f(a[i..i+len-1], b, h, 1)  
      return answer  
      
    f(s[1..len], b[1..len], h, index)  
      if index = len+1 then  
        return 1  
      for i = 1 to len  
        if s[index] + b[i] >= h  
          mem = b[i]  
          b[i] = 0  
          res = f(s, b, h, index + 1)  
          b[i] = mem  
          if res > 0  
            return 1  
      return 0  
    

Your task is to help Manao optimize his algorithm.

Input

The first line contains space-separated integers n, len and h (1 ≤ len ≤ n ≤ 150000; 1 ≤ h ≤ 109). The second line contains len space-separated integers b1, b2, ..., blen (1 ≤ bi ≤ 109). The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single number — the answer to Manao's problem.

Examples

Input

5 2 10
5 3
1 8 5 5 7


Output

2
#include <bits/stdc++.h>
using namespace std;
int n, m, K, v[150005], a[150005];
int q[600005], t[600005];
void Add(int k, int l, int r, int x, int d) {
  if (!x) return;
  if (x >= r) {
    q[k] += d;
    t[k] += d;
    return;
  }
  if (t[k])
    t[k << 1] += t[k], q[k << 1] += t[k], t[(k << 1) ^ 1] += t[k],
        q[(k << 1) ^ 1] += t[k], t[k] = 0;
  int m = l + r >> 1;
  if (x > m) Add((k << 1) ^ 1, m + 1, r, x, d);
  Add(k << 1, l, m, x, d);
  q[k] = min(q[k << 1], q[(k << 1) ^ 1]);
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1, x; i <= m; i++) scanf("%d", &x), v[i] = K - x;
  sort(v + 1, v + 1 + m);
  for (int i = 1, x; i <= n; i++)
    scanf("%d", &x), a[i] = upper_bound(v + 1, v + 1 + m, x) - v - 1;
  for (int i = 1; i <= m; i++) Add(1, 1, m, i, -1), Add(1, 1, m, a[i], 1);
  int ans = q[1] == 0;
  for (int i = m + 1; i <= n; i++) {
    Add(1, 1, m, a[i - m], -1);
    Add(1, 1, m, a[i], 1);
    ans += q[1] == 0;
  }
  printf("%d\n", ans);
}
