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 = 150003;
int n, len, h;
int b[MAXN], a[MAXN];
void read() {
  cin >> n >> len >> h;
  for (int i = 1; i <= len; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> a[i];
}
int tr[MAXN * 4], lazy[MAXN * 4];
void merge(int v) { tr[v] = min(tr[v * 2], tr[v * 2 + 1]); }
void build(int v, int l, int r) {
  if (l == r) {
    tr[v] = -len + l - 1;
    return;
  }
  int mid = (l + r) / 2;
  build(v * 2, l, mid);
  build(v * 2 + 1, mid + 1, r);
  merge(v);
}
void push(int v, int l, int r) {
  tr[v] += lazy[v];
  if (l != r) lazy[v * 2] += lazy[v], lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void update(int v, int l, int r, int ll, int rr, int val) {
  push(v, l, r);
  if (l > rr || r < ll) return;
  if (l >= ll && r <= rr) {
    lazy[v] += val;
    push(v, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(v * 2, l, mid, ll, rr, val);
  update(v * 2 + 1, mid + 1, r, ll, rr, val);
  merge(v);
}
vector<int> diff;
int find_idx(int val) {
  auto it = upper_bound(diff.begin(), diff.end(), val);
  if (it == diff.begin()) return -1;
  return it - diff.begin();
}
void solve() {
  for (int i = 1; i <= len; i++) diff.push_back(h - b[i]);
  sort(diff.begin(), diff.end());
  build(1, 1, len);
  for (int i = 1; i <= len; i++) {
    int curr_idx = find_idx(a[i]);
    if (curr_idx != -1) update(1, 1, len, 1, curr_idx, 1);
  }
  int ans = 0;
  if (tr[1] >= 0) ans++;
  for (int i = len + 1; i <= n; i++) {
    int curr_idx = find_idx(a[i]);
    if (curr_idx != -1) update(1, 1, len, 1, curr_idx, 1);
    curr_idx = find_idx(a[i - len]);
    if (curr_idx != -1) update(1, 1, len, 1, curr_idx, -1);
    if (tr[1] >= 0) ans++;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}
