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;
int mn[150000 * 4], flag[150000 * 4];
void push_down(int u) {
  mn[(u << 1)] += flag[u];
  mn[((u << 1) + 1)] += flag[u];
  flag[(u << 1)] += flag[u];
  flag[((u << 1) + 1)] += flag[u];
  flag[u] = 0;
}
void build(int u, int L, int R, int N) {
  if (R == L + 1) {
    mn[u] = L - N;
    return;
  }
  build((u << 1), L, ((L + R) >> 1), N);
  build(((u << 1) + 1), ((L + R) >> 1), R, N);
  mn[u] = min(mn[(u << 1)], mn[((u << 1) + 1)]);
}
void update(int u, int L, int R, int l, int r, int x) {
  l = max(L, l);
  r = min(R, r);
  if (l >= r) return;
  if (l == L && R == r) {
    flag[u] += x;
    mn[u] += x;
    return;
  }
  push_down(u);
  update((u << 1), L, ((L + R) >> 1), l, r, x);
  update(((u << 1) + 1), ((L + R) >> 1), R, l, r, x);
  mn[u] = min(mn[(u << 1)], mn[((u << 1) + 1)]);
}
int N, M, H;
int b[150000], a[150000];
int main() {
  scanf("%d%d%d", &N, &M, &H);
  for (int i = 0; i < M; i++) scanf("%d", &b[i]);
  for (int i = 0; i < N; i++) scanf("%d", &a[i]);
  for (int i = 0; i < M; i++) b[i] = H - b[i];
  sort(b, b + M);
  int ans = 0;
  build(1, 0, M, M);
  for (int i = 0, j = 0; i <= N - M; i++) {
    while (j < i + M) {
      int k = upper_bound(b, b + M, a[j]) - b;
      if (k > 0) update(1, 0, M, 0, k, 1);
      j++;
    }
    if (mn[1] >= 0) ans++;
    int k = upper_bound(b, b + M, a[i]) - b;
    if (k > 0) update(1, 0, M, 0, k, -1);
  }
  printf("%d\n", ans);
  return 0;
}
