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 INF = 1000 * 1000 * 1000;
const double PI = acos(-1.0);
const int MAX = 150005;
int n, len, h;
pair<int, int> a[MAX];
int b[MAX];
pair<int, int> T[MAX * 8];
int lazy[MAX * 8];
bool pzdc[MAX];
void push(int v) {
  T[v].first += lazy[v];
  lazy[v * 2] += lazy[v];
  lazy[v * 2 + 1] += lazy[v];
  lazy[v] = 0;
}
void upd(int v, int tl, int tr, int l, int r, int delta) {
  push(v);
  if (l > r) return;
  if (tl == l && tr == r) {
    lazy[v] += delta;
    push(v);
    return;
  }
  int tm = (tl + tr) / 2;
  upd(v * 2, tl, tm, l, min(r, tm), delta);
  upd(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, delta);
  T[v] = min(T[v * 2], T[v * 2 + 1]);
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    T[v].second = tl;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  T[v] = min(T[v * 2], T[v * 2 + 1]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> len >> h;
  build(1, 0, n - len);
  for (int i = 0; i < len; i++) cin >> b[i];
  sort(b, b + len);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a, a + n);
  int ptr = n - 1;
  for (int i = 0; i < len; i++) {
    while (ptr >= 0 && a[ptr].first + b[i] >= h) {
      upd(1, 0, n - len, max(0, a[ptr].second - len + 1),
          min(n - len, a[ptr].second), 1);
      ptr--;
    }
    while (T[1].first <= i) {
      pzdc[T[1].second] = 1;
      upd(1, 0, n - len, T[1].second, T[1].second, INF);
    }
  }
  int ans = 0;
  for (int i = 0; i < n - len + 1; i++) ans += !pzdc[i];
  cout << ans;
}
