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;
const long long maxn = 15e4 + 7;
long long f[maxn * 8], full[maxn * 8], n, len, h, a[maxn], b[maxn], ans;
vector<long long> v;
map<long long, long long> m;
void upd(long long x, long long l, long long r, long long ll, long long val) {
  if (l > ll || r < ll) return;
  if (l == r) {
    f[x] += val;
    full[x] += val;
    return;
  }
  long long md = (l + r) / 2;
  if (ll <= md)
    upd(x * 2 + 0, l, md, ll, val);
  else
    upd(x * 2 + 1, md + 1, r, ll, val);
  full[x] = full[x * 2] + full[x * 2 + 1];
  f[x] = max(f[x * 2], full[x * 2] + f[x * 2 + 1]);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> len >> h;
  for (long long i = 0; i < len; ++i) {
    cin >> b[i];
    v.push_back(h - b[i]);
  }
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  long long cur = 1;
  for (long long i = 0; i < v.size(); ++i)
    if (i == 0 || v[i] != v[i - 1]) m[v[i]] = cur++;
  for (long long i = 0; i < len; ++i) upd(1, 1, cur - 1, m[h - b[i]], -1);
  for (long long i = 0; i < len; ++i) upd(1, 1, cur - 1, m[a[i]], 1);
  for (long long i = len; i <= n; ++i) {
    if (f[1] == 0) ++ans;
    if (i < n) {
      upd(1, 1, cur - 1, m[a[i - len]], -1);
      upd(1, 1, cur - 1, m[a[i]], 1);
    }
  }
  cout << ans << endl;
}
