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 N = 2e5 + 9;
vector<int> key;
vector<int>::iterator it;
int a[N], st[N * 4], lazy[N * 4], len;
void update(int l, int r, int x, int val, int m) {
  int mid = l + r >> 1;
  if (lazy[m]) {
    st[m] += lazy[m];
    if (l < r) {
      lazy[m * 2] += lazy[m];
      lazy[m * 2 + 1] += lazy[m];
    }
    lazy[m] = 0;
  }
  if (l > x) return;
  if (r <= x) {
    st[m] += val;
    if (l < r) {
      lazy[m * 2] += val;
      lazy[m * 2 + 1] += val;
    }
    return;
  }
  update(l, mid, x, val, m * 2);
  update(mid + 1, r, x, val, m * 2 + 1);
  st[m] = max(st[m * 2], st[m * 2 + 1]);
}
void construct(int l, int r, int m) {
  int mid = l + r >> 1;
  if (l == r) {
    st[m] = len - l + 1;
    return;
  }
  construct(l, mid, m * 2);
  construct(mid + 1, r, m * 2 + 1);
  st[m] = max(st[m * 2], st[m * 2 + 1]);
}
int main() {
  int n, h, x, ans = 0, i, j;
  scanf("%d %d %d", &n, &len, &h);
  for (i = 1; i <= len; i++) {
    scanf("%d", &x);
    key.push_back(h - x);
  }
  construct(1, len, 1);
  sort(key.begin(), key.end());
  for (i = 1; i <= len; i++) {
    scanf("%d", &a[i]);
    it = upper_bound(key.begin(), key.end(), a[i]);
    it--;
    j = it - key.begin() + 1;
    update(1, len, j, -1, 1);
  }
  if (st[1] <= 0) ans++;
  for (i = len + 1; i <= n; i++) {
    scanf("%d", &a[i]);
    it = upper_bound(key.begin(), key.end(), a[i]);
    it--;
    j = it - key.begin() + 1;
    update(1, len, j, -1, 1);
    it = upper_bound(key.begin(), key.end(), a[i - len]);
    it--;
    j = it - key.begin() + 1;
    update(1, len, j, 1, 1);
    if (st[1] <= 0) {
      ans++;
    }
  }
  printf("%d", ans);
}
