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>
const int N = 300005, M = 15000000, v = 1000000000, lim = 2000000000;
int n, m, h, tot, rt, a[N], b[N], ls[M], rs[M], mi[M], tag[M];
void add(int u, int x) {
  mi[u] += x;
  tag[u] += x;
}
void update(int &u, int l, int r, int ql, int qr, int x) {
  if (!u) u = ++tot;
  if (l >= ql && r <= qr) {
    add(u, x);
    return;
  }
  if (tag[u]) {
    if (!ls[u]) ls[u] = ++tot;
    add(ls[u], tag[u]);
    if (!rs[u]) rs[u] = ++tot;
    add(rs[u], tag[u]);
    tag[u] = 0;
  }
  int mid = l + (r - l) / 2;
  if (ql <= mid) update(ls[u], l, mid, ql, qr, x);
  if (qr > mid) update(rs[u], mid + 1, r, ql, qr, x);
  mi[u] = mi[ls[u]];
  if (mi[rs[u]] < mi[u]) mi[u] = mi[rs[u]];
}
int main() {
  scanf("%d%d%d", &n, &m, &h);
  for (int i = 1; i <= m; i++)
    scanf("%d", b + i), update(rt, 0, lim, b[i] + v, lim, -1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    update(rt, 0, lim, h - a[i] + v, lim, 1);
    if (i < m) continue;
    if (mi[rt] >= 0) ans++;
    update(rt, 0, lim, h - a[i - m + 1] + v, lim, -1);
  }
  printf("%d\n", ans);
}
