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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const int N = 2e5 + 4, MX = -(1 << 30);
int n, len, h, a[N], b[N], tree[2 * N + 400], lazy[2 * N + 400], lowbound[N],
    wynik;
void insert(int v, int l, int r, int a, int b, int c) {
  if (lazy[v] != 0) {
    tree[v] += lazy[v];
    if (r - l + 1 > 1) {
      lazy[2 * v] += lazy[v];
      lazy[2 * v + 1] += lazy[v];
    }
    lazy[v] = 0;
  }
  if (r < a || l > b || l > r) return;
  if (a <= l && r <= b) {
    tree[v] += c;
    if (r - l + 1 > 1) {
      lazy[2 * v] += c;
      lazy[2 * v + 1] += c;
    }
    return;
  }
  int mid = (l + r) / 2;
  insert(2 * v, l, mid, a, b, c);
  insert(2 * v + 1, mid + 1, r, a, b, c);
  tree[v] = max(tree[2 * v], tree[2 * v + 1]);
}
void build(int v, int l, int r) {
  if (l > r) return;
  if (l == r) {
    tree[v] = -(len - l + 1);
    return;
  }
  int mid = (l + r) / 2;
  build(2 * v, l, mid);
  build(2 * v + 1, mid + 1, r);
  tree[v] = max(tree[2 * v], tree[2 * v + 1]);
}
int main() {
  scanf("%d%d%d", &n, &len, &h);
  for (int i = 1; i <= len; i++) scanf("%d", &b[i]);
  sort(b + 1, b + (len + 1));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    lowbound[i] = lower_bound(b + 1, b + (len + 1), h - a[i]) - b;
    if (lowbound[i] == len + 1) lowbound[i] = 0;
  }
  build(1, 1, len);
  int pom = 0;
  for (int i = 1; i <= len; i++) {
    insert(1, 1, len, 1, lowbound[i] - 1, 1);
    if (lowbound[i] == 0) pom++;
  }
  for (int i = 1; i <= n - len + 1; i++) {
    if (tree[1] + lazy[1] < 0 && pom == 0) wynik++;
    insert(1, 1, len, 1, lowbound[i] - 1, -1);
    if (lowbound[i] == 0) pom--;
    insert(1, 1, len, 1, lowbound[i + len] - 1, 1);
    if (lowbound[i + len] == 0) pom++;
  }
  printf("%d\n", wynik);
  getchar();
  getchar();
  return 0;
}
