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 pp {
  int lc, rc, s, k;
} p[5000010];
int root, pl;
int A[150010], B[150010];
void build(int &x, int l, int r, int k, int s) {
  if (!x) x = ++pl;
  p[x].s += s;
  if (l == r) {
    p[x].k += s;
    return;
  }
  int mid = (l + r) >> 1;
  if (k <= mid)
    build(p[x].lc, l, mid, k, s);
  else
    build(p[x].rc, mid + 1, r, k, s);
  p[x].k = min(p[p[x].lc].k, p[p[x].lc].s + p[p[x].rc].k);
}
int main() {
  int n, m, h, x, i, j, ans = 0;
  scanf("%d%d%d", &n, &m, &h);
  for (i = 1; i <= m; i++) {
    scanf("%d", &x);
    build(root, 1, 1000000000, max(h - x, 1), 1);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    build(root, 1, 1000000000, A[i], -1);
    if (i >= m + 1) build(root, 1, 1000000000, A[i - m], 1);
    if (i >= m && p[root].k >= 0) ans++;
  }
  printf("%d\n", ans);
  return 0;
}
