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 MAXN = 2e5 + 47, inf = 1e17 + 47;
struct node {
  long long l, r, add, mn;
  node(long long p = 0) {
    l = r = -1;
    add = 0;
    mn = inf;
  }
};
long long n, len, H;
vector<node> t(4 * MAXN);
void build(long long i, long long l, long long r) {
  t[i].l = l, t[i].r = r;
  if (l == r) {
    t[i].add = 0, t[i].mn = len - l - 1;
    return;
  }
  long long mid = (l + r) / 2;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  t[i].mn = min(t[i * 2].mn, t[i * 2 + 1].mn);
}
void push(long long i) {
  t[i].mn += t[i].add;
  if (t[i].l != t[i].r) {
    t[i * 2].add += t[i].add;
    t[i * 2 + 1].add += t[i].add;
  }
  t[i].add = 0;
}
void change(long long i, long long l, long long r, long long c) {
  push(i);
  if (r < t[i].l || t[i].r < l) return;
  if (l <= t[i].l && t[i].r <= r) {
    t[i].add += c;
    push(i);
    return;
  }
  change(i * 2, l, r, c);
  change(i * 2 + 1, l, r, c);
  t[i].mn = min(t[i * 2].mn, t[i * 2 + 1].mn);
}
long long get_min() {
  push(1);
  return t[1].mn;
}
void print(long long i) {
  push(i);
  if (t[i].l == t[i].r)
    cout << t[i].mn << " ";
  else
    print(i * 2), print(i * 2 + 1);
}
void read() {
  cin >> n >> len >> H;
  vector<long long> a(n), b(len);
  for (long long i = 0; i < len; i++) cin >> b[i];
  for (long long i = 0; i < n; i++) cin >> a[i];
  build(1, 0, len - 1);
  sort(b.begin(), b.end());
  long long r = 0, ans = 0;
  for (long long l = 0; l < n; l++) {
    if (l + len > n) continue;
    while (r < l + len) {
      long long ind = (lower_bound(b.begin(), b.end(), H - a[r]) - b.begin());
      if (ind < len) change(1, ind, len - 1, 1);
      r++;
    }
    if (get_min() >= len) ans++;
    long long ind = (lower_bound(b.begin(), b.end(), H - a[l]) - b.begin());
    if (ind < len) change(1, ind, len - 1, -1);
  }
  cout << ans << "\n";
}
void solve() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    read();
    solve();
  }
}
