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;
const int maxn = 300000 + 10;
const int nn = 65536 * 8;
int n, len, h;
int a[maxn], b[maxn];
int best[maxn];
int tree[nn * 2 + 10], tag[nn * 2 + 10];
int fix;
void _down(int id) {
  tree[id * 2] += tag[id], tag[id * 2] += tag[id];
  tree[id * 2 + 1] += tag[id], tag[id * 2 + 1] += tag[id];
  tag[id] = 0;
}
void _update(int a, int b, int l, int r, int id) {
  if (a > r or b < l) return;
  if (l >= a and r <= b) {
    tree[id] += fix;
    tag[id] += fix;
    return;
  }
  _down(id);
  _update(a, b, l, (l + r) / 2, id * 2);
  _update(a, b, (l + r) / 2 + 1, r, id * 2 + 1);
  tree[id] = max(tree[id * 2], tree[id * 2 + 1]);
}
int _query(int a, int b, int l, int r, int id) {
  if (a > r or b < l) return -1e9;
  if (l >= a and r <= b) return tree[id];
  _down(id);
  return max(_query(a, b, l, (l + r) / 2, id * 2),
             _query(a, b, (l + r) / 2 + 1, r, id * 2 + 1));
}
set<int> ms;
int ANS;
void check() {
  if (_query(1, len, 1, nn, 1) <= 0 and ms.size() == 0) ANS++;
}
int main() {
  scanf("%d%d%d", &n, &len, &h);
  for (int i = 1; i <= len; i++) scanf("%d", &b[i]), b[i] = h - b[i];
  sort(b + 1, b + len + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    int l = 1, r = len, res = 0;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (a[i] >= b[mid])
        res = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    best[i] = res;
  }
  for (int i = 1; i <= len; i++) fix = -i, _update(i, i, 1, nn, 1);
  for (int i = 1; i <= len; i++) {
    fix = 1;
    if (best[i] == 0)
      ms.insert(i);
    else
      _update(best[i], len, 1, nn, 1);
  }
  check();
  for (int i = 1; i <= n - len; i++) {
    int now = i;
    fix = -1;
    if (best[now] == 0)
      ms.erase(now);
    else
      _update(best[now], len, 1, nn, 1);
    now = i + len - 1 + 1;
    fix = 1;
    if (best[now] == 0)
      ms.insert(now);
    else
      _update(best[now], len, 1, nn, 1);
    check();
  }
  cout << ANS << "\n";
  return 0;
}
