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 Maxn = 600005;
int minx[Maxn], tip[Maxn];
int a[Maxn], b[Maxn], c[Maxn];
int n, m, i, h, L, R, t, ans;
void ins(int p, int l, int r, int pos, int data) {
  if (l == r) {
    minx[p] = data;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    ins(p * 2 + 1, l, mid, pos, data);
  else
    ins(p * 2 + 2, mid + 1, r, pos, data);
  minx[p] = min(minx[p * 2 + 1], minx[p * 2 + 2]);
}
void push(int p) {
  minx[p * 2 + 1] += tip[p];
  tip[p * 2 + 1] += tip[p];
  minx[p * 2 + 2] += tip[p];
  tip[p * 2 + 2] += tip[p];
  tip[p] = 0;
}
void add(int p, int l, int r, int L, int R, int dlt) {
  if (L > r || l > R) return;
  if (L <= l && R >= r) {
    minx[p] += dlt;
    tip[p] += dlt;
    return;
  }
  if (tip[p]) push(p);
  int mid = (l + r) >> 1;
  add(p * 2 + 1, l, mid, L, R, dlt);
  add(p * 2 + 2, mid + 1, r, L, R, dlt);
  minx[p] = min(minx[p * 2 + 1], minx[p * 2 + 2]);
}
int main() {
  scanf("%d%d%d", &n, &m, &h);
  for (i = 1; i <= m; i++) scanf("%d", &b[i]);
  sort(b + 1, b + m + 1);
  for (i = 1; i <= m / 2; i++) swap(b[i], b[m - i + 1]);
  for (i = 1; i <= m; i++) b[i] = h - b[i];
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= m; i++) c[i] = a[i];
  sort(c + 1, c + m + 1);
  for (i = 1; i <= m; i++) {
    t = lower_bound(c + 1, c + m + 1, b[i]) - c;
    ins(0, 1, m, i, i - t);
  }
  ans += (minx[0] >= 0);
  for (i = m + 1; i <= n; i++) {
    if (a[i] > a[i - m]) {
      L = upper_bound(b + 1, b + m + 1, a[i - m]) - b;
      R = upper_bound(b + 1, b + m + 1, a[i]) - b - 1;
      if (L <= R && L <= m) add(0, 1, m, L, R, 1);
    } else if (a[i] < a[i - m]) {
      L = upper_bound(b + 1, b + m + 1, a[i]) - b;
      R = upper_bound(b + 1, b + m + 1, a[i - m]) - b - 1;
      if (L <= R && L <= m) add(0, 1, m, L, R, -1);
    }
    ans += (minx[0] >= 0);
  }
  printf("%d\n", ans);
  return 0;
}
