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 = 150000 + 23;
int n, len, h, seg[MAXN << 2], mj[MAXN << 2], a[MAXN], b[MAXN], ans;
void add(int l, int r, int val, int st = 0, int ed = len, int v = 1) {
  if (l >= ed || st >= r) return;
  if (l <= st && ed <= r) {
    mj[v] += val;
    return;
  }
  int mid = (st + ed) >> 1;
  add(l, r, val, st, mid, v << 1), add(l, r, val, mid, ed, (v << 1) | 1);
  seg[v] = min(seg[v << 1] + mj[v << 1], seg[(v << 1) | 1] + mj[(v << 1) | 1]);
}
int main() {
  cin >> n >> len >> h;
  for (int i = 0; i < len; i++) {
    cin >> b[i];
    b[i] = h - b[i];
  }
  sort(b, b + len);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = upper_bound(b, b + len, a[i]) - b;
  }
  for (int i = 1; i <= len; i++) add(0, i, -1);
  for (int i = 0; i < len - 1; i++) add(0, a[i], 1);
  for (int i = len - 1; i < n; i++) {
    add(0, a[i], 1);
    if (seg[1] + mj[1] >= 0) ans++;
    add(0, a[i - len + 1], -1);
  }
  cout << ans;
  return 0;
}
