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;
long long int n, len, h, c;
vector<long long int> a, b;
long long int tree[(long long int)(1e6) + 5 << 2];
long long int lazy[(long long int)(1e6) + 5 << 2];
void update(long long int wai, long long int start, long long int end,
            long long int l, long long int r, long long int val) {
  if (lazy[wai] != 0) {
    tree[wai] += lazy[wai];
    if (start != end)
      lazy[wai * 2] += lazy[wai], lazy[wai * 2 + 1] += lazy[wai];
    lazy[wai] = 0;
  }
  if (start > r || end < l) return;
  if (start >= l && end <= r) {
    tree[wai] += val;
    if (start != end) lazy[wai * 2] += val, lazy[wai * 2 + 1] += val;
    return;
  }
  update(wai * 2, start, start + (end - start) / 2, l, r, val),
      update(wai * 2 + 1, start + (end - start) / 2 + 1, end, l, r, val);
  tree[wai] = min(tree[wai * 2], tree[wai * 2 + 1]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> len >> h;
  for (long long int i = 0; i < len; i++) {
    cin >> c;
    b.push_back(h - c);
  }
  sort(b.begin(), b.end());
  for (long long int i = 0; i < n; i++) {
    cin >> c;
    a.push_back(c);
    a[i] = upper_bound(b.begin(), b.end(), a[i]) - b.begin();
  }
  for (long long int i = 1; i <= len; i++) update(1, 1, len, i, i, i - len - 1);
  for (long long int i = 0; i < len; i++) update(1, 1, len, 1, a[i], 1);
  long long int ans = 0;
  if (tree[1] > -1) ans++;
  for (long long int i = 1; i < n - len + 1; i++) {
    update(1, 1, len, 1, a[i - 1], -1);
    update(1, 1, len, 1, a[i + len - 1], 1);
    if (tree[1] > -1) ans++;
  }
  cout << ans << '\n';
}
