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 a[200010], b[200010], addmark[600010], minn[600010], tot = 0;
void pushdown(int rt) {
  if (addmark[rt]) {
    addmark[rt * 2] += addmark[rt];
    addmark[rt * 2 + 1] += addmark[rt];
    minn[rt * 2] += addmark[rt];
    minn[rt * 2 + 1] += addmark[rt];
    addmark[rt] = 0;
  }
}
void update(int rt, int l, int r, int x, int y, int z) {
  if (l > y || r < x) return;
  if (l >= x && r <= y) return addmark[rt] += z, minn[rt] += z, void();
  pushdown(rt);
  int mid = (l + r) / 2;
  update(rt * 2, l, mid, x, y, z);
  update(rt * 2 + 1, mid + 1, r, x, y, z);
  minn[rt] = min(minn[rt * 2], minn[rt * 2 + 1]);
}
int main() {
  int n, m, h, t, ans = 0;
  cin >> n >> m >> h;
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) update(1, 1, m, i, i, -i);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= m; i++)
    update(1, 1, m, lower_bound(a + 1, a + m + 1, h - b[i]) - a, m, 1);
  ans += (minn[1] >= 0);
  for (int i = m + 1; i <= n; i++) {
    update(1, 1, m, lower_bound(a + 1, a + m + 1, h - b[i - m]) - a, m, -1);
    update(1, 1, m, lower_bound(a + 1, a + m + 1, h - b[i]) - a, m, 1);
    ans += (minn[1] >= 0);
  }
  cout << ans << endl;
  return 0;
}
