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 A[150010];
int B[150010];
struct val_type {
  int value;
  int mn_val;
} VAL[2 * (1 << 18)];
void addit(int x, int A, int B, int a, int b, int v) {
  if (a <= A && B <= b) {
    VAL[x].value += v;
    VAL[x].mn_val += v;
    return;
  } else if (b <= A || B <= a) {
    return;
  }
  int M = (A + B) / 2;
  int c1 = 2 * x + 1;
  int c2 = c1 + 1;
  addit(c1, A, M, a, b, v);
  addit(c2, M, B, a, b, v);
  VAL[x].mn_val = VAL[x].value + min(VAL[c1].mn_val, VAL[c2].mn_val);
}
int main() {
  int N, len, H;
  cin >> N >> len >> H;
  for (int i = 0; i < len; i++) {
    cin >> B[i];
  }
  sort(B, B + len);
  for (int i = 0; i < len; i++) {
    addit(0, 0, (1 << 18), i, (1 << 18), -1);
  }
  int result = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A[i] = lower_bound(B, B + len, H - A[i]) - B;
    addit(0, 0, (1 << 18), A[i], (1 << 18), 1);
    if (i >= len) {
      addit(0, 0, (1 << 18), A[i - len], (1 << 18), -1);
    }
    result += VAL[0].mn_val >= 0;
  }
  cout << result << endl;
  return 0;
}
