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 = 2000010;
int mx[N], add[N], a[N], b[N];
void modify(int x, int l, int r, int ll, int rr, int v) {
  if (l > rr || ll > r) return;
  if (l >= ll && r <= rr) {
    add[x] += v;
    return;
  }
  if (add[x] != 0) {
    add[x + x] += add[x];
    add[x + x + 1] += add[x];
    add[x] = 0;
  }
  int y = (l + r) >> 1;
  modify(x + x, l, y, ll, rr, v);
  modify(x + x + 1, y + 1, r, ll, rr, v);
  mx[x] = mx[x + x + 1] + add[x + x + 1];
  if (mx[x + x] + add[x + x] > mx[x]) mx[x] = mx[x + x] + add[x + x];
}
int main() {
  int n, len, h;
  scanf("%d %d %d", &n, &len, &h);
  for (int i = 1; i <= len; i++) scanf("%d", b + i);
  sort(b + 1, b + len + 1);
  reverse(b + 1, b + len + 1);
  for (int i = 1; i <= n; i++) {
    int foo;
    scanf("%d", &foo);
    int ll = 0, rr = len;
    while (ll < rr) {
      int mid = (ll + rr + 1) >> 1;
      if (foo + b[mid] >= h)
        ll = mid;
      else
        rr = mid - 1;
    }
    a[i] = ll;
  }
  memset(mx, 0, sizeof(mx));
  memset(add, 0, sizeof(add));
  for (int i = 1; i <= len; i++) modify(1, 0, len, i, i, -i);
  for (int i = 1; i <= len - 1; i++) modify(1, 0, len, a[i], len, 1);
  int ans = 0;
  for (int i = len; i <= n; i++) {
    modify(1, 0, len, a[i], len, 1);
    if (mx[1] + add[1] == 0) ans++;
    modify(1, 0, len, a[i - len + 1], len, -1);
  }
  printf("%d\n", ans);
  return 0;
}
