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;
const int N = 150 * 1000 + 20;
int n, len, a[N], b[N], h, idx[N], seg[N << 2], lazy[N << 2], ans;
void shift(int id) {
  seg[id << 1] += lazy[id];
  seg[id << 1 | 1] += lazy[id];
  lazy[id << 1] += lazy[id];
  lazy[id << 1 | 1] += lazy[id];
  lazy[id] = 0;
}
void build(int id = 1, int st = 0, int en = len) {
  if (en - st == 1) {
    seg[id] = len - st - 1;
    return;
  }
  int mid = st + en >> 1;
  build(id << 1, st, mid);
  build(id << 1 | 1, mid, en);
  seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
}
void update(int l, int r, int val, int id = 1, int st = 0, int en = len) {
  if (r <= st || en <= l) return;
  if (l <= st && en <= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  int mid = st + en >> 1;
  shift(id);
  update(l, r, val, id << 1, st, mid);
  update(l, r, val, id << 1 | 1, mid, en);
  seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> len >> h;
  for (int i = 0; i < len; i++) {
    cin >> b[i];
    b[i] = min(b[i], h);
  }
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(b, b + len);
  b[len++] = h;
  for (int i = 0; i < n; i++)
    idx[i] = lower_bound(b, b + len, h - a[i]) - b + 1;
  build();
  for (int i = 0; i < len - 1; i++) update(0, idx[i], -1);
  ans = seg[1] >= 0;
  for (int i = len - 1; i < n; i++) {
    update(0, idx[i - len + 1], 1);
    update(0, idx[i], -1);
    ans += seg[1] >= 0;
  }
  return cout << ans << "\n", 0;
}
