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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int mod = (int)1e9 + 7;
int n, m, k, BASE;
int add[150010 * 4], mx[150010 * 4], b[150010], a[150010];
void get(int x, int l, int r) {
  if (l > r) return;
  int L = BASE + l - 1, R = BASE + r + 1;
  for (; (L ^ R) != 1; L /= 2, R /= 2) {
    if (~L & 1) add[L ^ 1] += x;
    if (R & 1) add[R ^ 1] += x;
    mx[L / 2] = min(add[L] + mx[L], add[L ^ 1] + mx[L ^ 1]);
    mx[R / 2] = min(add[R] + mx[R], add[R ^ 1] + mx[R ^ 1]);
  }
  for (; L > 1; L /= 2) {
    mx[L / 2] = min(add[L] + mx[L], add[L ^ 1] + mx[L ^ 1]);
  }
}
void In() {
  scanf("%d%d%d", &m, &n, &k);
  ;
  for (int i = (1); i <= (n); i++) {
    int x;
    scanf("%d", &x);
    ;
    b[i] = k - x;
  }
  sort(b + 1, b + n + 1);
  BASE = 1;
  while (BASE < n + 2) BASE += BASE;
  for (int i = (1); i <= (n); i++) add[BASE + i] = i - 1;
  for (int i = (BASE + n) / 2; i; i--)
    mx[i] = min(mx[i + i] + add[i + i], mx[i + i + 1] + add[i + i + 1]);
  for (int i = 0; i < (m); i++) {
    int x;
    scanf("%d", &x);
    ;
    a[i] = lower_bound(b + 1, b + n + 1, x + 1) - b - 1;
  }
  int ans = 0;
  for (int i = 0; i < (m); i++) {
    get(-1, a[i] + 1, n);
    if (i >= n) get(1, a[i - n] + 1, n);
    if (i >= n - 1) ans += ((mx[1] + add[1]) == 0);
  }
  cout << ans;
}
int main() {
  int _ = 1;
  while (_--) In();
  return 0;
}
