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;
int N, M, H, a[150005], b[150005], t[150005 * 4], d[150005 * 4];
void init() {
  scanf("%d%d%d", &N, &M, &H);
  for (int i = 1; i <= M; i++) scanf("%d", &b[i]), b[i] = H - b[i];
  sort(b + 1, b + M + 1);
  for (int i = 1, x; i <= N; i++)
    scanf("%d", &x), a[i] = upper_bound(b + 1, b + M + 1, x) - b - 1;
}
void build(int p, int L, int R) {
  if (L == R) {
    t[p] = -L;
    return;
  }
  build((p << 1), L, ((L + R) >> 1)),
      build(((p << 1) + 1), ((L + R) >> 1) + 1, R),
      t[p] = max(t[(p << 1)], t[((p << 1) + 1)]);
}
void add(int p, int L, int R, int l, int r, int x) {
  if (l <= L && R <= r) {
    d[p] += x, t[p] += x;
    return;
  }
  if (d[p])
    d[(p << 1)] += d[p], t[(p << 1)] += d[p], d[((p << 1) + 1)] += d[p],
        t[((p << 1) + 1)] += d[p], d[p] = 0;
  if (r <= ((L + R) >> 1))
    add((p << 1), L, ((L + R) >> 1), l, r, x);
  else if (((L + R) >> 1) < l)
    add(((p << 1) + 1), ((L + R) >> 1) + 1, R, l, r, x);
  else
    add((p << 1), L, ((L + R) >> 1), l, ((L + R) >> 1), x),
        add(((p << 1) + 1), ((L + R) >> 1) + 1, R, ((L + R) >> 1) + 1, r, x);
  t[p] = max(t[(p << 1)], t[((p << 1) + 1)]);
}
void doit() {
  build(1, 0, N);
  int ans = 0;
  for (int i = 1; i <= N; i++) {
    add(1, 0, N, a[i], N, 1);
    if (i > M) add(1, 0, N, a[i - M], N, -1);
    if (i >= M) ans += t[1] <= 0;
  }
  printf("%d\n", ans);
}
int main() {
  init();
  doit();
  return 0;
}
