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;
struct seg {
  struct treedot {
    int l, r;
    int sum, add;
    void update(int x) {
      sum += x;
      add += x;
    }
  } c[800005];
  void pushdown(int id) {
    if (c[id].add) {
      c[id * 2].update(c[id].add);
      c[id * 2 + 1].update(c[id].add);
      c[id].add = 0;
    }
  }
  void pushup(int id) { c[id].sum = min(c[id * 2].sum, c[id * 2 + 1].sum); }
  void build(int id, int l, int r) {
    if (l > r) return;
    c[id].l = l;
    c[id].r = r;
    c[id].sum = 0;
    c[id].add = 0;
    if (l == r)
      c[id].sum = 0;
    else {
      int mid = l + (r - l) / 2;
      build(id * 2, l, mid);
      build(id * 2 + 1, mid + 1, r);
      pushup(id);
    }
  }
  void update(int id, int le, int ri, int x) {
    if (le > c[id].r || ri < c[id].l) return;
    if (le <= c[id].l && c[id].r <= ri)
      c[id].update(x);
    else {
      pushdown(id);
      update(id * 2, le, ri, x);
      update(id * 2 + 1, le, ri, x);
      pushup(id);
    }
  }
};
seg tr;
int n, m, k, i, j, x, b[150005], a[150005], ans;
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    b[i] = k - b[i];
  }
  sort(b + 1, b + m + 1);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  tr.build(1, 1, m);
  for (i = 1; i <= m; i++) tr.update(1, i, i, i - m - 1);
  for (i = 1; i <= n; i++) {
    a[i] = upper_bound(b + 1, b + m + 1, a[i]) - b - 1;
  }
  for (i = 1; i <= m; i++) {
    tr.update(1, 1, a[i], 1);
  }
  ans += (tr.c[1].sum >= 0);
  for (i = m + 1; i <= n; i++) {
    tr.update(1, 1, a[i - m], -1);
    tr.update(1, 1, a[i], 1);
    ans += (tr.c[1].sum >= 0);
  }
  cout << ans << endl;
  return 0;
}
