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 maxn = 15e4 + 5;
int seg[4 * maxn], lazy[4 * maxn], a[maxn], b[maxn];
void relax(int id) {
  int x = lazy[id];
  lazy[id] = 0;
  seg[2 * id + 0] += x;
  lazy[2 * id + 0] += x;
  seg[2 * id + 1] += x;
  lazy[2 * id + 1] += x;
}
void add(int id, int L, int R, int l, int r, int val) {
  if (R <= l || r <= L) return;
  if (L >= l && R <= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  int mid = (L + R) / 2;
  relax(id);
  add(2 * id + 0, L, mid, l, r, val);
  add(2 * id + 1, mid, R, l, r, val);
  seg[id] = max(seg[2 * id], seg[2 * id + 1]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, len, h;
  cin >> n >> len >> h;
  for (int i = 0; i < len; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < len; i++) {
    b[i] = h - b[i];
  }
  sort(b, b + len);
  for (int i = 0; i < n; i++) {
    a[i] = upper_bound(b, b + len, a[i]) - b;
  }
  for (int i = 0; i < len; i++) {
    add(1, 0, len + 1, i + 1, len + 1, -1);
    add(1, 0, len + 1, a[i], len + 1, +1);
  }
  int ans = 0;
  if (seg[1] <= 0) ans++;
  for (int i = len; i < n; i++) {
    add(1, 0, len + 1, a[i - len], len + 1, -1);
    add(1, 0, len + 1, a[i], len + 1, +1);
    if (seg[1] <= 0) ans++;
  }
  cout << ans;
}
