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;
vector<int> v;
int mini[4000010];
int marked[4000010];
int b[150010];
void shift(int v) {
  if (marked[v] != 0) {
    marked[2 * v] += marked[v];
    marked[2 * v + 1] += marked[v];
    mini[2 * v] += marked[v];
    mini[2 * v + 1] += marked[v];
    marked[v] = 0;
  }
}
void add(int v, int f, int e, int l, int r, int k) {
  shift(v);
  if (f >= r || l >= e) return;
  if (f >= l && e <= r) {
    mini[v] += k;
    marked[v] += k;
    return;
  }
  int mid = (f + e) / 2;
  add(2 * v, f, mid, l, min(mid, r), k);
  add(2 * v + 1, mid, e, max(mid, l), r, k);
  mini[v] = min(mini[2 * v], mini[2 * v + 1]);
}
void insert(int k) {
  int p = upper_bound(v.begin(), v.end(), k) - v.begin();
  if (v[p - 1] < k) p++;
  add(1, 0, 500000, 0, p, -1);
}
void erase(int k) {
  int p = upper_bound(v.begin(), v.end(), k) - v.begin();
  if (v[p - 1] < k) p++;
  add(1, 0, 500000, 0, p, 1);
}
int main() {
  int n, len, h;
  cin >> n >> len >> h;
  v.push_back(-1000000001);
  for (int i = 0; i < len; i++) {
    int k;
    cin >> k;
    v.push_back(k);
  }
  sort(v.begin(), v.end());
  v.push_back(v.back() + 1);
  for (int i = 1; i <= len; i++) erase(v[i]);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (i >= len) erase(h - b[i - len]);
    insert(h - b[i]);
    if (i >= len - 1 && mini[1] == 0) {
      ans++;
    }
  }
  cout << ans << endl;
}
