Manao is solving a problem with the following statement:

<image>

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;
template <class T>
T cross(complex<T> a, complex<T> b) {
  return imag(a * conj(b));
}
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
void setmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
void setmax(T &a, T b) {
  if (b > a) a = b;
}
const int MAXN = 250000;
const int MAXL = 18, S = 1 << (MAXL + 1);
const int MAXM = MAXN;
int N, M;
int first[MAXN], second[MAXM];
int H;
int size[S];
int mval[S];
int lazy[S];
void init() {
  for (int i = S - 1; i > 0; i--) {
    if (i * 2 >= S) {
      size[i] = 1;
      mval[i] = -(M - (i - S / 2));
      lazy[i] = 0;
    } else {
      size[i] = size[2 * i] + size[2 * i + 1];
      mval[i] = min(mval[2 * i], mval[2 * i + 1]);
      lazy[i] = 0;
    }
  }
}
void update(int a, int diff, int i = 1) {
  if (a == 0) return;
  if (size[i] == 1) {
    lazy[i] += diff;
    mval[i] += lazy[i];
    lazy[i] = 0;
    return;
  }
  lazy[2 * i] += lazy[i], lazy[2 * i + 1] += lazy[i];
  lazy[i] = 0;
  int s = size[2 * i];
  if (a >= s) {
    update(a - s, diff, 2 * i + 1);
    lazy[2 * i] += diff;
  } else {
    update(a, diff, 2 * i);
  }
  mval[i] = min(mval[2 * i] + lazy[2 * i], mval[2 * i + 1] + lazy[2 * i + 1]);
}
int go() {
  int cnt = 0;
  sort(second, second + M);
  init();
  for (int i = 0; i < N; i++) {
    first[i] = second + M - lower_bound(second, second + M, H - first[i]);
  }
  for (int i = 0; i < M; i++) {
    update(first[i], 1);
  }
  if (mval[1] >= 0) cnt++;
  for (int i = M; i < N; i++) {
    update(first[i - M], -1);
    update(first[i], 1);
    if (mval[1] >= 0) cnt++;
  }
  return cnt;
}
void io(istream &in, ostream &out) {
  in >> N >> M >> H;
  for (int i = 0; i < M; i++) in >> second[i];
  for (int i = 0; i < N; i++) in >> first[i];
  out << go() << '\n';
}
int main() {
  ifstream fin("338E.in");
  if (fin.good()) {
    ofstream fout("338E.out");
    io(fin, fout);
    fin.close();
    fout.close();
  } else
    io(cin, cout);
  return 0;
}
