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 maxn = (int)3e5 + 1;
int n, m, h;
int a[maxn], b[maxn];
vector<int> v;
int t[maxn * 4], add[maxn * 4];
void push(int v, int tl, int tr) {
  if (tl != tr) {
    add[v * 2] += add[v];
    add[v * 2 + 1] += add[v];
  }
  t[v] += add[v];
  add[v] = 0;
}
void upd1(int v, int tl, int tr, int p, int q) {
  if (tl == tr) {
    t[v] = q;
  } else {
    int mid = (tl + tr) / 2;
    if (p <= mid) {
      upd1(v * 2, tl, mid, p, q);
    } else {
      upd1(v * 2 + 1, mid + 1, tr, p, q);
    }
    t[v] = max(t[v * 2], t[v * 2 + 1]);
  }
}
void upd(int v, int tl, int tr, int l, int r, int h) {
  push(v, tl, tr);
  if (l > tr || tl > r) {
    return;
  }
  if (l <= tl && tr <= r) {
    add[v] = h;
    push(v, tl, tr);
    return;
  }
  int mid = (tl + tr) / 2;
  upd(v * 2, tl, mid, l, r, h);
  upd(v * 2 + 1, mid + 1, tr, l, r, h);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
int ans;
int main() {
  cin >> n >> m >> h;
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    v.push_back(h - a[i]);
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    v.push_back(b[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 1; i <= n; i++) {
    b[i] = lower_bound(v.begin(), v.end(), b[i]) - v.begin() + 1;
  }
  for (int i = 1; i <= m; i++) {
    a[i] = lower_bound(v.begin(), v.end(), h - a[i]) - v.begin() + 1;
  }
  sort(a + 1, a + m + 1);
  reverse(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) {
    upd1(1, 1, 300000, a[i], i);
  }
  for (int i = 1; i <= m; i++) {
    upd(1, 1, 300000, 1, b[i], -1);
  }
  if (t[1] < 1) {
    ans++;
  }
  for (int i = m + 1; i <= n; i++) {
    upd(1, 1, 300000, 1, b[i - m], 1);
    upd(1, 1, 300000, 1, b[i], -1);
    if (t[1] < 1) {
      ans++;
    }
  }
  cout << ans;
}
