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;
long long n, len, h, ans, a[400005], b[400005], c[400005], lazy[400005];
void pushdown(long long x, long long l, long long r) {
  long long mid = l + r >> 1;
  c[x << 1] += lazy[x];
  lazy[x << 1] += lazy[x];
  c[x << 1 | 1] += lazy[x];
  lazy[x << 1 | 1] += lazy[x];
  lazy[x] = 0;
}
void update(long long x, long long l, long long r, long long ll, long long rr,
            long long s) {
  if (l >= ll && r <= rr) {
    c[x] += s;
    lazy[x] += s;
    return;
  }
  pushdown(x, l, r);
  long long mid = l + r >> 1;
  if (ll <= mid) update(x << 1, l, mid, ll, rr, s);
  if (rr > mid) update(x << 1 | 1, mid + 1, r, ll, rr, s);
  c[x] = min(c[x << 1], c[x << 1 | 1]);
}
signed main() {
  scanf("%lld%lld%lld", &n, &len, &h);
  for (long long i = 1; i <= len; i++) scanf("%lld", &b[i]);
  sort(b + 1, b + 1 + len);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] = lower_bound(b + 1, b + 1 + len, h - a[i]) - b;
  }
  for (long long i = 1; i <= len; i++) update(1, 1, len, i, i, -i);
  for (long long i = 1; i < len; i++) update(1, 1, len, a[i], len, 1);
  for (long long i = len; i <= n; i++) {
    update(1, 1, len, a[i], len, 1);
    if (c[1] >= 0) ans++;
    update(1, 1, len, a[i - len + 1], len, -1);
  }
  printf("%lld\n", ans);
  return 0;
}
