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 mx[2000010], add[2000010], a[2000010], b[2000010];
void update(int p, int l, int r, int x, int y, int v) {
  if (x <= l && y >= r) {
    add[p] += v;
    return;
  }
  int m = (l + r) >> 1;
  if (add[p]) {
    add[p << 1] += add[p];
    add[p << 1 | 1] += add[p];
    add[p] = 0;
  }
  if (x <= m) update(p << 1, l, m, x, y, v);
  if (y > m) update(p << 1 | 1, m + 1, r, x, y, v);
  mx[p] = max(mx[p << 1] + add[p << 1], mx[p << 1 | 1] + add[p << 1 | 1]);
}
int main() {
  int i, j, k, n, m, len, h;
  while (~scanf("%d%d%d", &n, &len, &h)) {
    for (i = 1; i <= len; i++) {
      scanf("%d", &b[i]);
    }
    sort(b + 1, b + len + 1);
    reverse(b + 1, b + len + 1);
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      int l = 0, r = len;
      while (l < r) {
        m = (l + r + 1) >> 1;
        if (a[i] + b[m] >= h)
          l = m;
        else
          r = m - 1;
      }
      a[i] = l;
    }
    memset(mx, 0, sizeof(mx));
    memset(add, 0, sizeof(add));
    for (i = 1; i <= len; i++) update(1, 0, len, i, i, -i);
    for (i = 1; i < len; i++) update(1, 0, len, a[i], len, 1);
    int ans = 0;
    for (i = len; i <= n; i++) {
      update(1, 0, len, a[i], len, 1);
      if (mx[1] + add[1] == 0) ans++;
      update(1, 0, len, a[i - len + 1], len, -1);
    }
    printf("%d\n", ans);
  }
  return 0;
}
