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 M = 1e9 + 7;
const int N = 2e5 + 7;
vector<int> b;
int a[N];
int n, m, h;
int seg[N * 4];
int laz[N * 4];
void spl(int v) {
  for (int u = (v * 2); u <= (v * 2 + 1); u++) {
    seg[u] += laz[v];
    laz[u] += laz[v];
  }
  laz[v] = 0;
}
void add(int l, int r, int val, int v = 1, int s = 0, int e = b.size()) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    seg[v] += val;
    laz[v] += val;
    return;
  }
  spl(v);
  add(l, r, val, (v * 2), s, ((s + e) / 2));
  add(l, r, val, (v * 2 + 1), ((s + e) / 2), e);
  seg[v] = min(seg[(v * 2)], seg[(v * 2 + 1)]);
}
int sml(int x) { return upper_bound(b.begin(), b.end(), x) - b.begin(); }
int main() {
  cin >> n >> m >> h;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    b.push_back(h - x);
  }
  sort(b.begin(), b.end());
  for (int i = 1; i <= n; i++) cin >> a[i];
  int res = 0;
  for (int i = m - 1; i >= 0; i--) add(i, i + 1, i - m);
  for (int i = 1; i <= m; i++) add(0, sml(a[i]), 1);
  if (seg[1] >= 0) res++;
  for (int i = m + 1; i <= n; i++) {
    add(0, sml(a[i]), 1);
    add(0, sml(a[i - m]), -1);
    if (seg[1] >= 0) res++;
  }
  cout << res;
}
