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;
template <size_t t_Size>
struct Segment_Tree {
  int tr[4 * t_Size], lazy[4 * t_Size];
  Segment_Tree() {}
  void build(int index, int l, int r, const int &len) {
    if (l == r) {
      tr[index] = l - len;
      return;
    }
    int mid = (l + r) >> 1;
    build(2 * index + 1, l, mid, len);
    build(2 * index + 2, mid + 1, r, len);
    tr[index] = min(tr[2 * index + 1], tr[2 * index + 2]);
  }
  void push_lazy(int index, int l, int r) {
    if (lazy[index]) {
      tr[index] += lazy[index];
      if (l != r) {
        lazy[2 * index + 1] += lazy[index];
        lazy[2 * index + 2] += lazy[index];
      }
      lazy[index] = 0;
    }
  }
  void update(int index, int l, int r, int sl, int sr, int value) {
    push_lazy(index, l, r);
    if (r < sl || l > sr) return;
    if (sl <= l && r <= sr) {
      lazy[index] = value;
      push_lazy(index, l, r);
      return;
    }
    int mid = (l + r) >> 1;
    update(2 * index + 1, l, mid, sl, sr, value);
    update(2 * index + 2, mid + 1, r, sl, sr, value);
    tr[index] = min(tr[2 * index + 1], tr[2 * index + 2]);
  }
  int query(int index, int l, int r, int sl, int sr) {
    push_lazy(index, l, r);
    if (r < sl || l > sr) return 0;
    if (sl <= l && r <= sr) return tr[index];
    int mid = (l + r) >> 1;
    int lvalue = query(2 * index + 1, l, mid, sl, sr);
    int rvalue = query(2 * index + 2, mid + 1, r, sl, sr);
    return min(lvalue, rvalue);
  }
};
const int k_N = 15e4 + 3;
int a[k_N], b[k_N];
int n, len, h;
Segment_Tree<k_N> segment_tree;
int get_index(int x) { return (upper_bound(b, b + len, x) - b) - 1; }
void solve() {
  for (int i = 0; i < len; ++i) b[i] = h - b[i];
  sort(b, b + len);
  segment_tree.build(0, 0, len - 1, len);
  for (int i = 0; i < len; ++i) {
    int index = get_index(a[i]);
    if (index != -1) segment_tree.update(0, 0, len - 1, 0, index, 1);
  }
  int answer = 0;
  if (segment_tree.tr[0] >= 0) answer++;
  for (int i = len; i < n; ++i) {
    int index = get_index(a[i]);
    if (index != -1) segment_tree.update(0, 0, len - 1, 0, index, 1);
    index = get_index(a[i - len]);
    if (index != -1) segment_tree.update(0, 0, len - 1, 0, index, -1);
    if (segment_tree.tr[0] >= 0) answer++;
  }
  cout << answer << "\n";
}
void read() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> len >> h;
  for (int i = 0; i < len; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> a[i];
}
int main() {
  read();
  solve();
}
