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;
struct tree {
  int add;
  int mn;
  int l;
  int r;
};
tree tr[500000];
void init(int x, int l, int r) {
  tr[x].add = 0;
  tr[x].l = l;
  tr[x].r = r;
  if (l == r) {
    tr[x].mn = -l - 1;
    return;
  }
  init(2 * x, l, (l + r) / 2);
  init(2 * x + 1, (l + r) / 2 + 1, r);
  tr[x].mn = min(tr[2 * x].mn, tr[2 * x + 1].mn);
}
void add(int x, int l, int val) {
  if (tr[x].r < l) return;
  if (tr[x].l >= l) {
    tr[x].add += val;
    return;
  }
  add(2 * x, l, val);
  add(2 * x + 1, l, val);
  tr[x].mn =
      min(tr[2 * x].mn + tr[2 * x].add, tr[2 * x + 1].mn + tr[2 * x + 1].add);
}
int n, len, h;
int a[150005], b[150005];
int main() {
  cin >> n >> len >> h;
  int i, j;
  for (i = 0; i < len; i++) scanf("%d", &b[i]);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(b, b + len);
  init(1, 0, len - 1);
  int tb[n];
  int ans = 0;
  for (i = 0; i < n; i++) {
    int l = -1;
    int r = len;
    while (l < r - 1) {
      if (a[i] + b[(l + r) / 2] >= h)
        r = (l + r) / 2;
      else
        l = (l + r) / 2;
    }
    tb[i] = r;
    add(1, r, 1);
    if (i >= len) add(1, tb[i - len], -1);
    if (tr[1].mn + tr[1].add >= 0) ++ans;
  }
  printf("%d\n", ans);
}
