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;
const int N = 200020, D = N << 2;
int n, m, h, a[N], b[N], p[N], ans;
int bad, mx[D], ad[D];
void build(int d, int lt, int rt) {
  if (lt == rt) {
    mx[d] = -lt;
    return;
  }
  int md = (lt + rt) >> 1;
  build(d << 1 | 0, lt, md);
  build(d << 1 | 1, md + 1, rt);
  mx[d] = -lt;
}
void mark(int d, int x) {
  ad[d] += x;
  mx[d] += x;
}
void push(int d) {
  if (ad[d]) {
    mark(d << 1 | 0, ad[d]);
    mark(d << 1 | 1, ad[d]);
    ad[d] = 0;
  }
}
void add(int d, int lt, int rt, int lq, int rq, int x) {
  if (lq <= lt && rt <= rq) {
    mark(d, x);
    return;
  }
  push(d);
  int md = (lt + rt) >> 1;
  if (lq <= md) add(d << 1 | 0, lt, md, lq, rq, x);
  if (md < rq) add(d << 1 | 1, md + 1, rt, lq, rq, x);
  mx[d] = max(mx[d << 1 | 0], mx[d << 1 | 1]);
}
void add(int i, int x) {
  if (!p[i]) {
    bad += x;
    return;
  }
  add(1, 1, m, p[i], m, x);
}
int main() {
  scanf("%d%d%d", &n, &m, &h);
  for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(b + 1, b + m + 1);
  build(1, 1, m);
  for (int i = 1; i <= n; ++i) {
    p[i] = lower_bound(b + 1, b + m + 1, h - a[i]) - b;
    p[i] = m + 1 - p[i];
    add(i, 1);
    if (i > m) add(i - m, -1);
    ans += (!bad && i >= m && mx[1] <= 0);
  }
  printf("%d\n", ans);
}
