Manao is solving a problem with the following statement:

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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;
#pragma GCC optimize("Ofast")
const long long inf = 1e18 + 5;
const long long N = 2e5 + 5;
long long n, len, h, a[N], b[N], ans;
long long seg[N << 2], lazy[N << 2];
void build(long long id = 1, long long l = 0, long long r = N) {
  if (r - l == 1) {
    seg[id] = -l;
    return;
  }
  long long mid = l + r >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid, r);
  seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
}
void shift(long long id) {
  lazy[id << 1] += lazy[id];
  lazy[id << 1 | 1] += lazy[id];
  seg[id << 1] += lazy[id];
  seg[id << 1 | 1] += lazy[id];
  lazy[id] = 0;
}
void update(long long ql, long long qr, long long val, long long id = 1,
            long long l = 0, long long r = N) {
  if (qr <= l || ql >= r || qr <= ql || r <= l) return;
  if (ql <= l && qr >= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  long long mid = (l + r) / 2;
  shift(id);
  update(ql, qr, val, id << 1, l, mid);
  update(ql, qr, val, id << 1 | 1, mid, r);
  seg[id] = min(seg[id << 1], seg[id << 1 | 1]);
}
long long get(long long ql, long long qr, long long id = 1, long long l = 0,
              long long r = N) {
  if (qr <= l || ql >= r || qr <= ql || r <= l) return inf;
  if (ql <= l && qr >= r) return seg[id];
  long long mid = (l + r) / 2;
  shift(id);
  return min(get(ql, qr, id << 1, l, mid), get(ql, qr, id << 1 | 1, mid, r));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> len >> h;
  for (long long i = 1; i <= len; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build();
  sort(b + 1, b + len + 1);
  for (long long i = 1; i <= len; i++)
    update(lower_bound(b + 1, b + len + 1, h - a[i]) - b, len + 1, 1);
  long long ptr = 1;
  while (ptr + len - 1 <= n) {
    if (get(1, len + 1) >= 0) ans++;
    update(lower_bound(b + 1, b + len + 1, h - a[ptr]) - b, len + 1, -1);
    update(lower_bound(b + 1, b + len + 1, h - a[ptr + len]) - b, len + 1, 1);
    ptr++;
  }
  cout << ans << endl;
}
