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 = 150010;
int n, len, h;
int b[N], a[N];
int t[N * 4], s[N * 4];
void update(int node, int l, int r, int x, int y, int val) {
  if (x > r || y < l) {
    return;
  }
  if (x <= l && r <= y) {
    t[node] += val;
    s[node] += val;
    return;
  }
  int m = (l + r) >> 1;
  update(node + node, l, m, x, y, val);
  update(node + node + 1, m + 1, r, x, y, val);
  t[node] = max(t[node * 2], t[node * 2 + 1]) + s[node];
}
int main() {
  scanf("%d %d %d", &n, &len, &h);
  for (int i = 1; i <= len; i++) {
    scanf("%d", b + i);
  }
  sort(b + 1, b + len + 1, greater<int>());
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    int l = 0, r = n + 1;
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (u + b[m] >= h) {
        l = m;
      } else {
        r = m;
      }
    }
    a[i] = l;
  }
  for (int i = 0; i <= len; i++) {
    update(1, 0, len, i, i, -i);
  }
  for (int i = 1; i < len; i++) {
    update(1, 0, len, a[i], len, 1);
  }
  int res = 0;
  for (int i = len; i <= n; i++) {
    update(1, 0, len, a[i], len, 1);
    res += (t[1] == 0);
    update(1, 0, len, a[i - len + 1], len, -1);
  }
  cout << res << endl;
  return 0;
}
