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 N = 2e5, SQ = 390;
int n, len, h, ans, a[N], b[N], f[N], g[N];
struct SQ_PART {
  int cnt[SQ], lazy, sz, mini;
  SQ_PART(int X = SQ) { sz = X; }
  void change_sz(int X) {
    sz = X;
    for (int i = 0; i < sz; i++) mini = min(mini, cnt[i] - i);
  }
  void add(int l, int r, int val) {
    if (r - l == sz) {
      lazy += val;
      return;
    }
    for (; l < r; l++) cnt[l] += val;
    mini = N;
    for (int i = 0; i < sz; i++) mini = min(mini, cnt[i] - i);
  }
  int get() { return mini + lazy; }
} part[SQ];
void add(int l, int r, int val) {
  if (f[r - 1] == f[l])
    part[f[l]].add(g[l], g[r], val);
  else {
    part[f[l]].add(g[l], SQ, val);
    l = SQ * (f[l] + 1);
    add(l, r, val);
  }
}
void DEL(int x) { add(lower_bound(b, b + len, h - x) - b, len, -1); }
void ADD(int x) { add(lower_bound(b, b + len, h - x) - b, len, 1); }
bool check() {
  for (int i = 0; i * SQ < len; i++)
    if (part[i].get() <= i * SQ) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  for (int i = 0; i < N; i++) f[i] = i / SQ, g[i] = i % SQ;
  cin >> n >> len >> h;
  for (int i = 0; i * SQ < len; i++)
    if ((i + 1) * SQ > len) part[i].change_sz(len % SQ);
  for (int i = 0; i < len; i++) cin >> b[i];
  sort(b, b + len);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < len; i++) ADD(a[i]);
  ans += check();
  for (int i = len; i < n; i++) {
    DEL(a[i - len]);
    ADD(a[i]);
    ans += check();
  }
  cout << ans;
  return 0;
}
