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 mx = 2e5 + 5;
const int MOD = 1e9 + 7;
long long sz, ara[mx], aux[mx];
int mini[mx * 3], lazy[mx * 3];
void build(int pos, int L, int R) {
  if (L == R) {
    mini[pos] = sz - L;
    return;
  }
  int x = pos * 2, mid = (L + R) / 2;
  build(x, L, mid);
  build(x + 1, mid + 1, R);
  mini[pos] = min(mini[x], mini[x + 1]);
}
void propagate(int pos, int L, int R) {
  int x = pos * 2, mid = (L + R) / 2;
  mini[x] += lazy[pos];
  mini[x + 1] += lazy[pos];
  lazy[x] += lazy[pos];
  lazy[x + 1] += lazy[pos];
  lazy[pos] = 0;
}
void update(int pos, int L, int R, int l, int r, int val) {
  if (L > r || l > R) return;
  if (L >= l && R <= r) {
    mini[pos] += val;
    lazy[pos] += val;
    return;
  }
  if (lazy[pos] != 0) propagate(pos, L, R);
  int x = pos * 2, mid = (L + R) / 2;
  if (mid >= l) update(x, L, mid, l, r, val);
  if (mid < r) update(x + 1, mid + 1, R, l, r, val);
  mini[pos] = min(mini[x], mini[x + 1]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) cin >> aux[i];
  for (int i = 1; i <= n; i++) cin >> ara[i];
  sort(aux, aux + m);
  sz = m + 1;
  build(1, 1, sz);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    long long need = k - ara[i];
    int x = lower_bound(aux, aux + m, need) - aux + 1;
    update(1, 1, sz, 1, x, -1);
    if (i >= m) {
      if (mini[1] >= 0) ans++;
      long long auxneed = k - ara[i - m + 1];
      int z = lower_bound(aux, aux + m, auxneed) - aux + 1;
      update(1, 1, sz, 1, z, 1);
    }
  }
  cout << ans << endl;
  return 0;
}
