Manao is solving a problem with the following statement:

<image>

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 = 1.5e5 + 5;
int n, len, h, a[N], b[N], ans;
struct seg {
  int sum, tag;
} tr[N << 2];
void pushdown(int x) {
  tr[x << 1].tag += tr[x].tag;
  tr[x << 1 | 1].tag += tr[x].tag;
  tr[x].tag = 0;
}
void update(int l, int r, int x, int ql, int qr, int k) {
  if (l >= ql && r <= qr) {
    tr[x].tag += k;
    return;
  }
  if (tr[x].tag) pushdown(x);
  int mid = l + r >> 1;
  if (ql <= mid) update(l, mid, x << 1, ql, qr, k);
  if (qr > mid) update(mid + 1, r, x << 1 | 1, ql, qr, k);
  tr[x].sum = min(tr[x << 1].sum + tr[x << 1].tag,
                  tr[x << 1 | 1].sum + tr[x << 1 | 1].tag);
}
int main() {
  scanf("%d%d%d", &n, &len, &h);
  for (int i = 1; i <= len; ++i) scanf("%d", &b[i]);
  sort(b + 1, b + len + 1);
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i]), a[i] = lower_bound(b + 1, b + len + 1, h - a[i]) - b;
  for (int i = 1; i <= len; ++i) update(1, len, 1, i, i, -i);
  for (int i = 1; i < len; ++i) update(1, len, 1, a[i], len, 1);
  for (int i = len; i <= n; ++i) {
    update(1, len, 1, a[i], len, 1);
    if (tr[1].sum + tr[1].tag >= 0) ++ans;
    update(1, len, 1, a[i - len + 1], len, -1);
  }
  cout << ans;
  return 0;
}
