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;
int a[150000], b[150000], now[150000];
vector<int> v, le;
vector<pair<int, int> > p;
int check(int n, int h) {
  for (int i = 0; i < n; ++i)
    if (le[i] + now[n - i - 1] < h) return 0;
  return 1;
}
void add(int x, int n, int y = -1) {
  int i = n - 1;
  if (y >= 0) {
    for (i = 0; i < n; ++i)
      if (now[i] == y) break;
  }
  now[i] = x;
  while ((i > 0) && (now[i] < now[i - 1])) {
    swap(now[i], now[i - 1]);
    --i;
  }
  while ((i + 1 < n) && (now[i] > now[i + 1])) {
    swap(now[i], now[i + 1]);
    ++i;
  }
}
int main() {
  int n, m, h;
  while (scanf("%d %d %d ", &n, &m, &h) != EOF) {
    int ans = 0;
    for (int i = 0; i < m; ++i) scanf("%d ", &b[i]);
    for (int i = 0; i < n; ++i) {
      scanf("%d ", &a[i]);
      p.push_back(make_pair(a[i], i));
    }
    sort(p.begin(), p.end());
    int y = sqrt(m);
    y = min(m, 500);
    int x = m - y;
    sort(b, b + m);
    if (m <= 500) {
      le.clear();
      for (int i = 0; i < m; ++i) le.push_back(b[i]);
      for (int i = 0; i < m; ++i) add(a[i], i + 1);
      for (int i = m - 1; i < n; ++i) {
        ans += check(m, h);
        add(a[i + 1], m, a[i - m + 1]);
      }
      cout << ans << endl;
      continue;
    }
    for (int i = 0, j = 0, ok; i + m <= n; ++i) {
      if (i == j) {
        v.clear();
        le.clear();
        j = i + m - x + 1;
        for (int k = 0; k < y; ++k) add(a[i + k], k + 1);
        for (int k = 0; k < n; ++k)
          if ((p[k].second < i + m) && (p[k].second >= j - 1))
            v.push_back(p[k].first);
        int t = 0;
        ok = true;
        for (int k = v.size() - 1; k >= 0; --k) {
          while ((t < m) && (b[t] + v[k] < h)) {
            le.push_back(b[t]);
            ++t;
          }
          if (t == m) {
            ok = false;
            break;
          } else
            ++t;
        }
        while (t < m) {
          le.push_back(b[t]);
          ++t;
        }
      }
      if (ok) ans += check(y, h);
      add(a[i + m], y, a[i]);
    }
    cout << ans << endl;
  }
}
