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 M = 2e5 + 5, LG = 25, inf = 1e9, mod = 95542721;
int n, m, h;
int a[M], b[M];
int seg[3 * M], lzy[3 * M];
void build(int l = 0, int r = m, int id = 1) {
  if (l + 1 == r) {
    seg[id] = m - r;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, 2 * id);
  build(mid, r, 2 * id + 1);
  seg[id] = min(seg[2 * id], seg[2 * id + 1]);
}
void lzyupd(int id, int v) {
  seg[id] += v;
  lzy[id] += v;
}
void lzyref(int id) {
  lzyupd(2 * id, lzy[id]);
  lzyupd(2 * id + 1, lzy[id]);
  lzy[id] = 0;
}
void upd(int x, int y, int v, int l = 0, int r = m, int id = 1) {
  if (x <= l && r <= y) {
    lzyupd(id, v);
    return;
  }
  if (y <= l || r <= x) return;
  lzyref(id);
  int mid = (l + r) >> 1;
  upd(x, y, v, l, mid, 2 * id);
  upd(x, y, v, mid, r, 2 * id + 1);
  seg[id] = min(seg[2 * id], seg[2 * id + 1]);
}
void add(int x, int f) { upd(0, (lower_bound(b, (b) + (m), h - x) - (b)), f); }
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> h;
  build();
  for (int i = 0; i < m; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(b, b + m);
  for (int i = 0; i < m; i++) {
    add(a[i], -1);
  }
  int cnt = (seg[1] == 0);
  for (int i = m; i < n; i++) {
    add(a[i], -1);
    add(a[i - m], 1);
    cnt += (seg[1] == 0);
  }
  cout << cnt << '\n';
}
