Manao is solving a problem with the following statement:

<image>

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;
char _;
const int maxN = 150 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 9;
const int base = 701;
const int SQ = 317;
const int maxL = 17;
int a[maxN], b[maxN];
int seg[4 * maxN], ad[4 * maxN];
void shift(int ind) {
  ad[2 * ind + 0] += ad[ind];
  seg[2 * ind + 0] += ad[ind];
  ad[2 * ind + 1] += ad[ind];
  seg[2 * ind + 1] += ad[ind];
  ad[ind] = 0;
}
void add(int l, int r, int val, int s, int e, int ind) {
  if (l >= e || r <= s) return;
  if (l <= s && e <= r) {
    seg[ind] += val;
    ad[ind] += val;
    return;
  }
  shift(ind);
  int mid = (s + e) >> 1;
  add(l, r, val, s, mid, 2 * ind + 0);
  add(l, r, val, mid, e, 2 * ind + 1);
  seg[ind] = min(seg[2 * ind], seg[2 * ind + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, len, h;
  cin >> n >> len >> h;
  for (int i = 0; i < len; i++) add(i, i + 1, -i - 1, 0, len, 1);
  for (int i = 0; i < len; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(b, b + len);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int ps = lower_bound(b, b + len, h - a[i]) - b;
    add(ps, len, 1, 0, len, 1);
    if (i >= len) {
      ps = lower_bound(b, b + len, h - a[i - len]) - b;
      add(ps, len, -1, 0, len, 1);
    }
    if (i >= len - 1)
      if (seg[1] >= 0) ans++;
  }
  cout << ans << endl;
  return 0;
}
