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 = 2e5 + 100;
int n, m, h, ans, a[maxn], b[maxn];
int seg[4 * maxn], lzy[4 * maxn], f[maxn];
void add(int val, int l, int r, int L = 1, int R = m + 1, int id = 1) {
  if (l == L && r == R) {
    seg[id] += val;
    lzy[id] += val;
    return;
  }
  int mid = (L + R) >> 1;
  if (lzy[id] != 0) {
    add(lzy[id], L, mid, L, mid, id * 2 + 0);
    add(lzy[id], mid, R, mid, R, id * 2 + 1);
    lzy[id] = 0;
  }
  if (l < mid) add(val, l, min(mid, r), L, mid, id * 2 + 0);
  if (r > mid) add(val, max(l, mid), r, mid, R, id * 2 + 1);
  seg[id] = min(seg[id * 2 + 0], seg[id * 2 + 1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> h;
  for (int i = 0; i < m; i++) cin >> b[i];
  sort(b, b + m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) f[i] = m - (lower_bound(b, b + m, h - a[i]) - b);
  for (int i = 2; i <= m + 1; i++) add(-1, 1, i);
  for (int i = 0; i < m; i++)
    if (f[i] != 0) add(1, 1, f[i] + 1);
  if (seg[1] >= 0) ans++;
  for (int i = m; i < n; i++) {
    if (f[i] != 0) add(1, 1, f[i] + 1);
    if (f[i - m] != 0) add(-1, 1, f[i - m] + 1);
    if (seg[1] >= 0) ans++;
  }
  cout << ans << endl;
  return 0;
}
