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;
const int N = 15e4 + 4;
int n, m, a[N], b[N], t[4 * N], laz[4 * N], ans, h;
void update(int n, int b, int e, int u, int k) {
  if (u < b) return;
  if (e <= u) {
    laz[n] += k;
    return;
  }
  int l = n << 1, r = l | 1, m = b + e >> 1;
  update(l, b, m, u, k);
  update(r, m + 1, e, u, k);
  t[n] = max(t[l] + laz[l], t[r] + laz[r]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> h;
  for (int i = 0; i < m; i++) cin >> b[i];
  sort(b, b + m);
  for (int i = 0, tmp; i < n; i++) {
    cin >> tmp;
    a[i] = lower_bound(b, b + m, h - tmp) - b;
  }
  update(1, 0, m, m, m);
  for (int i = 0; i < m; i++) update(1, 0, m, i, -1);
  for (int i = 0; i < m - 1; i++) update(1, 0, m, a[i], 1);
  for (int i = m - 1; i < n; i++) {
    update(1, 0, m, a[i], 1);
    if (t[1] + laz[1] <= m) ans++;
    update(1, 0, m, a[i - m + 1], -1);
  }
  cout << ans << '\n';
  return 0;
}
