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 N = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
int a[N], b[N], mn[4 * N], lz[4 * N], n;
void build(int l = 0, int r = n, int id = 1) {
  if (r - l == 1) {
    mn[id] = -l;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * id);
  build(mid, r, 2 * id + 1);
  mn[id] = min(mn[2 * id], mn[2 * id + 1]);
}
void shift(int id) {
  int x = lz[id];
  mn[2 * id] += x;
  mn[2 * id + 1] += x;
  lz[2 * id] += x;
  lz[2 * id + 1] += x;
  lz[id] = 0;
}
void upd(int l, int r, int x, int s = 0, int e = n, int id = 1) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    mn[id] += x;
    lz[id] += x;
    return;
  }
  shift(id);
  int mid = (s + e) / 2;
  upd(l, r, x, s, mid, 2 * id);
  upd(l, r, x, mid, e, 2 * id + 1);
  mn[id] = min(mn[2 * id], mn[2 * id + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, h;
  cin >> m >> n >> h;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  b[0] = n;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    b[i] = lower_bound(a, a + n, h - x) - a;
  }
  int ans = 0;
  build();
  for (int i = 1; i < n; i++) upd(b[i], n, 1);
  for (int i = n; i <= m; i++) {
    upd(b[i - n], n, -1);
    upd(b[i], n, 1);
    if (1 <= mn[1]) ans++;
  }
  cout << ans;
  return 0;
}
