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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
int n, len, h, m;
int b[200005], ar[200005], val[200005];
struct segtree {
  int val[800005], lazy[800005];
  int n;
  void init(int n_) {
    n = 1;
    while (n < n_) n <<= 1;
    for (int i = 0; i < n * 2; ++i) val[i] = lazy[i] = 0;
  }
  void push(int i) {
    if (lazy[i]) {
      lazy[i * 2 + 1] += lazy[i];
      lazy[i * 2 + 2] += lazy[i];
      val[i * 2 + 1] += lazy[i];
      val[i * 2 + 2] += lazy[i];
      lazy[i] = 0;
    }
  }
  void add(int a, int b, int i, int l, int r, int v) {
    if (a <= l && r <= b) {
      lazy[i] += v;
      val[i] += v;
      return;
    }
    if (b <= l || r <= a) return;
    int md = (l + r) >> 1;
    push(i);
    add(a, b, i * 2 + 1, l, md, v);
    add(a, b, i * 2 + 2, md, r, v);
    val[i] = max(val[i * 2 + 1], val[i * 2 + 2]);
  }
  int query() { return val[0]; }
};
segtree seg;
int main() {
  cin >> n >> len >> h;
  for (int i = 0; i < len; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> ar[i];
  m = len;
  sort(b, b + m, greater<int>());
  for (int i = 0; i < m; ++i) b[i] = h - b[i];
  seg.init(m);
  for (int i = 0; i < n; ++i) ar[i] = upper_bound(b, b + m, ar[i]) - b;
  for (int i = 0; i < m; ++i) seg.add(i, i + 1, 0, 0, seg.n, m - i);
  int res = 0;
  for (int i = 0; i < m; ++i) seg.add(0, ar[i], 0, 0, seg.n, -1);
  if (seg.query() <= 0) ++res;
  for (int i = m; i < n; ++i) {
    int ad = ar[i], ers = ar[i - m];
    seg.add(0, ad, 0, 0, seg.n, -1);
    seg.add(0, ers, 0, 0, seg.n, 1);
    if (seg.query() <= 0) ++res;
  }
  cout << res << endl;
  return 0;
}
