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 = 2e5 + 7;
const int SS = 4 * N;
int mn[SS], add[SS];
int n, m, k, a[N], b[N];
void shift(int v) {
  mn[2 * v] += add[v];
  add[2 * v] += add[v];
  mn[2 * v + 1] += add[v];
  add[2 * v + 1] += add[v];
  add[v] = 0;
}
void update(int l, int r, int val, int s = 0, int e = n, int v = 1) {
  if (r <= s || l >= e) return;
  if (l <= s && e <= r) {
    add[v] += val;
    mn[v] += val;
    return;
  }
  shift(v);
  int mid = (s + e) / 2;
  update(l, r, val, s, mid, 2 * v);
  update(l, r, val, mid, e, 2 * v + 1);
  mn[v] = min(mn[2 * v], mn[2 * v + 1]);
}
int main() {
  scanf("%d %d %d", &m, &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < m; i++) scanf("%d", &a[i]);
  sort(b, b + n);
  for (int i = 0; i < n; i++) update(i, i + 1, -i - 1);
  for (int i = 0; i < m; i++) a[i] = lower_bound(b, b + n, k - a[i]) - b;
  for (int i = 0; i < n - 1; i++) update(a[i], n, 1);
  int ans = 0;
  for (int i = n - 1; i < m; i++) {
    update(a[i], n, 1);
    if (mn[1] >= 0) ans++;
    update(a[i - n + 1], n, -1);
  }
  printf("%d\n", ans);
}
