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>
const long long MAX_N = 5e5 + 10;
const long long MOD = 1e9 + 9;
using namespace std;
long long a[MAX_N];
long long b[MAX_N];
long long c[MAX_N];
long long mn[MAX_N];
long long fl[MAX_N];
long long sum[MAX_N];
long long n, m, h, y;
void shift(int k) {
  int l = k * 2;
  int r = l + 1;
  mn[l] += fl[k];
  mn[r] += fl[k];
  fl[l] += fl[k];
  fl[r] += fl[k];
  fl[k] = 0;
}
void update(int k) {
  int l = k * 2;
  int r = l + 1;
  mn[k] = min(mn[l], mn[r]);
}
void add(int l, int r, int a, int b, int k, int d) {
  if (r < a or b < l) return;
  if (a <= l and r <= b) {
    fl[k] += d;
    mn[k] += d;
    return;
  }
  shift(k);
  int mid = (l + r) / 2;
  add(l, mid, a, b, k * 2, d);
  add(mid + 1, r, a, b, k * 2 + 1, d);
  update(k);
}
void build(int k) {
  if (k >= y) return;
  int l = k * 2;
  int r = l + 1;
  build(l);
  build(r);
  update(k);
}
void solve(int i) {
  long long x = c[i];
  long long t = c[i - m];
  if (x != t) add(0, y - 1, min(x, t), max(x, t) - 1, 1, (t - x) / abs(x - t));
}
int main() {
  for (int i = 0; i < MAX_N; ++i) mn[i] = 1e9;
  cin >> n >> m >> h;
  y = (1 << ((int)ceil(log2(n))));
  for (int i = 0; i < m; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(b, b + m);
  for (int i = 0; i < n; ++i) {
    int l = -1;
    int r = m;
    while (r - l > 1) {
      int mid = (l + r) / 2;
      if (b[mid] + a[i] >= h)
        r = mid;
      else
        l = mid;
    }
    c[i] = r;
  }
  for (int i = 0; i < m; ++i) mn[i + y] = -(i + 1);
  long long ans = 0;
  for (int i = 0; i < m; ++i) sum[c[i]]++;
  for (int i = 0; i < m; ++i) {
    if (i) sum[i] += sum[i - 1];
    mn[i + y] += sum[i];
  }
  build(1);
  if (mn[1] >= 0) ans++;
  for (int i = m; i < n; ++i) {
    solve(i);
    if (mn[1] >= 0) ans++;
  }
  cout << ans;
  return 0;
}
