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 + 10;
int f[MAXN], lazy[4 * MAXN], seg[4 * MAXN], INF = 1e9;
int a[MAXN], b[MAXN];
void shift(int v) {
  lazy[2 * v] += lazy[v];
  lazy[2 * v + 1] += lazy[v];
  seg[2 * v] += lazy[v];
  seg[2 * v + 1] += lazy[v];
  lazy[v] = 0;
}
void upd(int v, int val, int l, int r, int s, int e) {
  if (l <= s and e <= r) {
    seg[v] = seg[v] + val;
    lazy[v] += val;
    return;
  }
  if (e <= l or r <= s) return;
  shift(v);
  int mid = (s + e) / 2;
  upd(2 * v, val, l, r, s, mid);
  upd(2 * v + 1, val, l, r, mid, e);
  seg[v] = min(seg[2 * v], seg[2 * v + 1]);
  return;
}
int get(int v, int l, int r, int s, int e) {
  if (l <= s and e <= r) return seg[v];
  if (e <= l or r <= s) return INF;
  shift(v);
  int mid = (s + e) / 2;
  return min(get(2 * v, l, r, s, mid), get(2 * v + 1, l, r, mid, e));
}
int main() {
  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++) upd(1, -i - 1, i, i + 1, 0, len);
  sort(b, b + len);
  for (int i = 0; i < n; i++) f[i] = lower_bound(b, b + len, h - a[i]) - b;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (f[i] < len) upd(1, +1, f[i], len, 0, len);
    if (i >= len and f[i - len] < len) upd(1, -1, f[i - len], len, 0, len);
    if (i >= len - 1) {
      int mn = get(1, 0, len, 0, len);
      if (mn >= 0) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
