Manao is solving a problem with the following statement:

<image>

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;
void read(int& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int N = 150010;
const int D = 600010;
int n, len, h, i, j, k, l, p, x, ans;
int b[N], a[N];
int key[D], add[D];
void padd(int i, int s) {
  key[i] += s;
  add[i] += s;
}
void pdown(int i) {
  if (add[i]) padd(i * 2, add[i]), padd(i * 2 + 1, add[i]), add[i] = 0;
}
void upd(int i) { key[i] = min(key[i * 2], key[i * 2 + 1]); }
void build(int i, int x, int y) {
  if (x != y) {
    build(i * 2, x, (x + y) / 2);
    build(i * 2 + 1, (x + y) / 2 + 1, y);
    upd(i);
  } else
    key[i] = x;
}
void find(int i, int x, int y, int l, int r, int s) {
  if (y < l || x > r) return;
  if (x >= l && y <= r) {
    padd(i, s);
    return;
  }
  pdown(i);
  find(i * 2, x, (x + y) / 2, l, r, s);
  find(i * 2 + 1, (x + y) / 2 + 1, y, l, r, s);
  upd(i);
}
int main() {
  read(n);
  read(len);
  read(h);
  for (i = 1; i <= len; i++) read(b[i]), b[i] = h - b[i];
  sort(b + 1, b + 1 + len);
  build(1, 0, n);
  for (i = 1; i <= n; i++) {
    read(x);
    p = upper_bound(b + 1, b + 1 + len, x) - b - 1;
    a[i] = p;
    find(1, 0, n, a[i], n, -1);
    if (i > len) find(1, 0, n, a[i - len], n, 1);
    if (i >= len) ans += (key[1] >= 0);
  }
  printf("%d\n", ans);
  scanf("\n");
  return 0;
}
