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 = 150 * 1000 + 10;
const long long INF = 10000000000000000;
int n, len, a[maxn], b[maxn], h, ans[4 * maxn], t[maxn], add[4 * maxn], jvb;
void bld(int x, int l, int r) {
  if (r - l == 1)
    ans[x] = -(l + 1);
  else {
    int mid = (l + r) / 2;
    bld(2 * x + 1, l, mid);
    bld(2 * x + 2, mid, r);
    ans[x] = ans[2 * x + 2];
  }
}
void push(int x, int l, int r) {
  ans[2 * x + 1] += add[x];
  ans[2 * x + 2] += add[x];
  add[2 * x + 1] += add[x];
  add[2 * x + 2] += add[x];
  add[x] = 0;
}
void update(int x, int l, int r, int ql, int qr, int val) {
  if (ql >= r || l >= qr) return;
  if (l >= ql && r <= qr) {
    ans[x] += val;
    add[x] += val;
    return;
  }
  int mid = (l + r) / 2;
  push(x, l, r);
  update(2 * x + 1, l, mid, ql, qr, val);
  update(2 * x + 2, mid, r, ql, qr, val);
  ans[x] = min(ans[2 * x + 1], ans[2 * x + 2]);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> len >> h;
  for (int i = 0; i < len; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(b, b + len);
  bld(0, 0, len);
  for (int i = 0; i < n; i++) {
    t[i] = lower_bound(b, b + len, h - a[i]) - b;
    if (i >= len) update(0, 0, len, t[i - len], len, -1);
    update(0, 0, len, t[i], len, 1);
    if (ans[0] >= 0) jvb++;
  }
  cout << jvb;
  return 0;
}
