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 = 1.5e5 + 10;
int n, m, h, a[maxn], b[maxn], minv[maxn << 2], laz[maxn << 2];
inline void pushdown(int root) {
  if (laz[root]) {
    laz[root << 1] += laz[root];
    laz[root << 1 | 1] += laz[root];
    minv[root << 1] += laz[root];
    minv[root << 1 | 1] += laz[root];
    laz[root] = 0;
  }
}
inline void pushup(int root) {
  minv[root] = min(minv[root << 1], minv[root << 1 | 1]);
}
inline void update(int ul, int ur, int x, int l = 1, int r = m, int root = 1) {
  if (l > ur || r < ul) return;
  if (ul <= l && r <= ur) return laz[root] += x, minv[root] += x, void();
  int mid = l + r >> 1;
  pushdown(root);
  update(ul, ur, x, l, mid, root << 1);
  update(ul, ur, x, mid + 1, r, root << 1 | 1);
  pushup(root);
}
int main() {
  scanf("%d%d%d", &n, &m, &h);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    update(i, m, -1);
  }
  sort(b + 1, b + m + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = lower_bound(b + 1, b + m + 1, h - a[i]) - b;
    if (i <= m) update(a[i], m, 1);
  }
  int ans = minv[1] >= 0;
  for (int i = m + 1; i <= n; i++) {
    update(a[i], m, 1);
    update(a[i - m], m, -1);
    ans += minv[1] >= 0;
  }
  printf("%d", ans);
  return 0;
}
