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 MAXN = 3e5 + 17;
const int INF = 1e9 + 17;
int n;
int seg[4 * MAXN], lazy[4 * MAXN];
int a[MAXN], b[MAXN];
void shift(int id) {
  seg[2 * id + 1] += lazy[id];
  lazy[2 * id + 1] += lazy[id];
  seg[2 * id + 2] += lazy[id];
  lazy[2 * id + 2] += lazy[id];
  lazy[id] = 0;
}
void update(int ql, int qr, int qv, int xl = 0, int xr = n, int id = 0) {
  if (ql == qr) return;
  if (ql <= xl && xr <= qr) {
    lazy[id] += qv;
    seg[id] += qv;
    return;
  }
  if (qr <= xl || xr <= ql) return;
  int mid = (xl + xr) / 2;
  shift(id);
  update(ql, qr, qv, xl, mid, 2 * id + 1);
  update(ql, qr, qv, mid, xr, 2 * id + 2);
  seg[id] = max(seg[2 * id + 1], seg[2 * id + 2]);
}
void build(int id = 0, int xl = 0, int xr = n) {
  if (xr - xl == 1) {
    seg[id] = xl + 1;
    return;
  }
  int mid = (xl + xr) / 2;
  build(2 * id + 1, xl, mid);
  build(2 * id + 2, mid, xr);
  seg[id] = max(seg[2 * id + 1], seg[2 * id + 2]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int k, h;
  cin >> k >> n >> h;
  build();
  for (int i = 0; i < n; i++) cin >> b[i];
  sort(b, b + n);
  for (int i = 0; i < k; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++)
    update(lower_bound(b, b + n, h - a[i]) - b, n, -1);
  int ans = 0;
  for (int i = n - 1; i < k; i++) {
    update(lower_bound(b, b + n, h - a[i]) - b, n, -1);
    ans += (seg[0] == 0);
    update(lower_bound(b, b + n, h - a[i - n + 1]) - b, n, 1);
  }
  cout << ans;
}
