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 long long N = 150009;
long long a[N];
long long b[N], n, m, h;
struct Tree {
  long long l, r;
  long long ly, mi;
} t[N << 2];
void build_tree(long long p, long long l, long long r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    t[p].mi = l - m - 1;
    t[p].ly = 0;
    return;
  }
  build_tree((p << 1), l, ((t[p].l + t[p].r) >> 1));
  build_tree(((p << 1) | 1), ((t[p].l + t[p].r) >> 1) + 1, r);
  t[p].mi = min(t[(p << 1)].mi, t[((p << 1) | 1)].mi);
}
void push_down(long long p) {
  if (t[p].ly) {
    t[(p << 1)].ly += t[p].ly;
    t[((p << 1) | 1)].ly += t[p].ly;
    t[(p << 1)].mi += t[p].ly;
    t[((p << 1) | 1)].mi += t[p].ly;
    t[p].ly = 0;
  }
}
void modify(long long p, long long l, long long r, long long v) {
  if (t[p].l == l && t[p].r == r) {
    t[p].mi += v;
    t[p].ly += v;
    return;
  }
  push_down(p);
  if (r <= ((t[p].l + t[p].r) >> 1))
    modify((p << 1), l, r, v);
  else if (l > ((t[p].l + t[p].r) >> 1))
    modify(((p << 1) | 1), l, r, v);
  else {
    modify((p << 1), l, ((t[p].l + t[p].r) >> 1), v);
    modify(((p << 1) | 1), ((t[p].l + t[p].r) >> 1) + 1, r, v);
  }
  t[p].mi = min(t[(p << 1)].mi, t[((p << 1) | 1)].mi);
}
long long query(long long p, long long l, long long r) {
  if (t[p].l == l && t[p].r == r) {
    return t[p].mi;
  }
  push_down(p);
  if (r <= ((t[p].l + t[p].r) >> 1))
    return query((p << 1), l, r);
  else if (l > ((t[p].l + t[p].r) >> 1))
    return query(((p << 1) | 1), l, r);
  else
    return min(query((p << 1), l, ((t[p].l + t[p].r) >> 1)),
               query(((p << 1) | 1), ((t[p].l + t[p].r) >> 1) + 1, r));
}
long long cal(long long x) {
  long long s = 1, t = m;
  while (s <= t) {
    long long i = (s + t) >> 1;
    if (b[i] > x)
      t = i - 1;
    else
      s = i + 1;
  }
  return t;
}
long long p[N];
int main() {
  cin >> n >> m >> h;
  for (long long i = 1; i <= m; i++) cin >> b[i], b[i] = h - b[i];
  sort(b + 1, b + m + 1);
  build_tree(1, 1, m);
  for (long long i = 0; i < m; i++) {
    long long v;
    cin >> v;
    p[i] = cal(v);
    if (p[i]) modify(1, 1, p[i], 1);
  }
  long long ans = (query(1, 1, m) >= 0);
  for (long long i = m; i < n; i++) {
    long long v;
    cin >> v;
    p[i] = cal(v);
    if (p[i]) modify(1, 1, p[i], 1);
    if (p[i - m]) modify(1, 1, p[i - m], -1);
    ans += (query(1, 1, m) >= 0);
  }
  cout << ans << endl;
}
