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 Read() {
  char c;
  while (c = getchar(), (c != '-') && (c < '0' || c > '9'))
    ;
  bool neg = (c == '-');
  int ret = (neg ? 0 : c - 48);
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + c - 48;
  return neg ? -ret : ret;
}
const int MAXN = 150005;
struct node {
  int d, m;
} t[1 << 19];
int N, M, H, ans, a[MAXN], b[MAXN];
void init() {
  scanf("%d%d%d", &N, &M, &H);
  for (int i = 1; i <= M; i++) b[i] = H - Read();
  sort(b + 1, b + M + 1);
  for (int i = 1; i <= N; i++)
    a[i] = Read(), a[i] = upper_bound(b + 1, b + M + 1, a[i]) - b - 1;
}
void Apply(int x, int v) { t[x].d += v, t[x].m += v; }
void Mod(int id, int l, int r, int x, int v) {
  if (!x) return;
  if (x >= r) {
    Apply(id, v);
    return;
  }
  if (t[id].d)
    Apply(id + id, t[id].d), Apply(id + id + 1, t[id].d), t[id].d = 0;
  int mid = (l + r) >> 1;
  Mod(id + id, l, mid, x, v);
  if (x > mid) Mod(id + id + 1, mid + 1, r, x, v);
  t[id].m = min(t[id + id].m, t[id + id + 1].m);
}
void work() {
  for (int i = 1; i <= M; i++) Mod(1, 1, M, i, -1), Mod(1, 1, M, a[i], 1);
  ans = (t[1].m == 0);
  for (int i = M + 1; i <= N; i++)
    Mod(1, 1, M, a[i - M], -1), Mod(1, 1, M, a[i], 1), ans += (t[1].m == 0);
  printf("%d\n", ans);
}
int main() {
  init();
  work();
  return 0;
}
