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;
const int N = 2e5 + 123;
const long long oo = 1e9 + 123;
int n, len;
long long h, a[N], b[N];
struct {
  int f[N * 4], lazy[N * 4];
  void build(int k, int l, int r) {
    if (l == r) {
      f[k] = l - len - 1;
    } else {
      int mid = (l + r) >> 1;
      build(k * 2, l, mid);
      build(k * 2 + 1, mid + 1, r);
      f[k] = min(f[k * 2], f[k * 2 + 1]);
    }
  }
  void down(int k, int l, int r) {
    f[k] += lazy[k];
    if (l < r) {
      lazy[k * 2] += lazy[k];
      lazy[k * 2 + 1] += lazy[k];
    }
    lazy[k] = 0;
  }
  void update(int k, int l, int r, int L, int R, int x) {
    down(k, l, r);
    if (L > r || l > R) return;
    if (L <= l && r <= R) {
      lazy[k] += x;
      down(k, l, r);
      return;
    }
    int mid = (l + r) >> 1;
    update(k * 2, l, mid, L, R, x);
    update(k * 2 + 1, mid + 1, r, L, R, x);
    f[k] = min(f[k * 2], f[k * 2 + 1]);
  }
} IT;
int getId(int x) { return upper_bound(b + 1, b + 1 + len, x) - 1 - b; }
void testcase() {
  cin >> n >> len >> h;
  for (int i = 1; i <= len; ++i) cin >> b[i];
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= len; ++i) b[i] = h - b[i];
  sort(b + 1, b + 1 + len);
  IT.build(1, 1, len);
  for (int i = 1; i <= len; ++i) {
    int j = getId(a[i]);
    if (j >= 1) IT.update(1, 1, len, 1, j, 1);
  }
  int ans = 0;
  if (IT.f[1] >= 0) {
    ++ans;
  }
  for (int i = len + 1; i <= n; ++i) {
    int j = getId(a[i]);
    if (j >= 1) IT.update(1, 1, len, 1, j, 1);
    j = getId(a[i - len]);
    if (j >= 1) IT.update(1, 1, len, 1, j, -1);
    if (IT.f[1] >= 0) ++ans;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int tcase = 1;
  for (int i = 1; i <= tcase; ++i) testcase();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
