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 int mxN = 150000;
int n, m, h, a[mxN], b[mxN], ans;
vector<int> d;
array<int, 2> st[4 * mxN];
array<int, 2> comb(array<int, 2> l, array<int, 2> r) {
  array<int, 2> res = {l[0] + r[0], l[1] + r[1]};
  int mn = min(l[0], r[1]);
  res[0] -= mn, res[1] -= mn;
  return res;
}
void upd(int i, array<int, 2> x, int u = 1, int l = 0, int r = d.size() - 1) {
  if (l == r) {
    st[u][0] += x[0];
    st[u][1] += x[1];
    int mn = min(st[u][0], st[u][1]);
    st[u][0] -= mn;
    st[u][1] -= mn;
    return;
  }
  int mid = (l + r) / 2;
  if (i <= mid)
    upd(i, x, 2 * u, l, mid);
  else
    upd(i, x, 2 * u + 1, mid + 1, r);
  st[u] = comb(st[2 * u], st[2 * u + 1]);
}
void dbg(int u = 1, int l = 0, int r = d.size() - 1) {
  cout << "[" << l << ", " << r << "] = [" << st[u][0] << ", " << st[u][1]
       << "]" << endl;
  if (l == r) return;
  int mid = (l + r) / 2;
  dbg(2 * u, l, mid);
  dbg(2 * u + 1, mid + 1, r);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> h;
  for (int i = 0; i < m; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> a[i];
  d = vector<int>(a, a + n);
  sort(d.begin(), d.end());
  d.resize(unique(d.begin(), d.end()) - d.begin());
  for (int i = 0; i < m; ++i) {
    int x = lower_bound(d.begin(), d.end(), h - b[i]) - d.begin();
    if (x == d.size()) {
      cout << 0;
      return 0;
    }
    upd(x, {1, 0});
  }
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(d.begin(), d.end(), a[i]) - d.begin();
    upd(a[i], {0, 1});
    if (i >= m) upd(a[i - m], {0, -1});
    ans += st[1][0] == 0;
  }
  cout << ans;
  return 0;
}
