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 maxn = 150 * 1000 + 100;
int n, m, h;
int b[maxn], a[maxn];
int min_val[4 * maxn], to_add[4 * maxn];
void add(int x, int l, int r, int p, int q, int v) {
  if (q <= l || r <= p) return;
  if (p <= l && r <= q) {
    min_val[x] += v;
    to_add[x] += v;
    return;
  }
  int md = (l + r) / 2;
  min_val[2 * x + 1] += to_add[x];
  to_add[2 * x + 1] += to_add[x];
  min_val[2 * x + 2] += to_add[x];
  to_add[2 * x + 2] += to_add[x];
  to_add[x] = 0;
  add(2 * x + 1, l, md, p, q, v);
  add(2 * x + 2, md, r, p, q, v);
  min_val[x] = min(min_val[2 * x + 1], min_val[2 * x + 2]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> m >> n >> h;
  for (int i = 0, _n = (int)(n); i < _n; i++) cin >> b[i], b[i] = h - b[i];
  sort(b, b + n);
  for (int i = 0, _n = (int)(m); i < _n; i++)
    cin >> a[i], a[i] = upper_bound(b, b + n, a[i]) - b;
  for (int i = 0, _n = (int)(n); i < _n; i++) add(0, 0, n, 0, i + 1, -1);
  int ans = 0;
  for (int i = 0, _n = (int)(m); i < _n; i++) {
    add(0, 0, n, 0, a[i], 1);
    if (i >= n) add(0, 0, n, 0, a[i - n], -1);
    if (min_val[0] >= 0) ans++;
  }
  cout << ans << endl;
  {
    int _;
    cin >> _;
    return 0;
  }
}
