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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int Maxn = 200010, Maxm = 300005, Mo = 1000000007,
          sp[8][2] = {{-1, 0}, {1, 0},   {0, 1},  {0, -1},
                      {1, 1},  {-1, -1}, {1, -1}, {-1, 1}};
const int oo = INT_MAX - 10000000;
typedef int IArr[Maxn * 3];
IArr a, b, g, t, to, mk;
pair<int, int> c[Maxn];
int n, m, i, j, k, h;
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second < b.second;
}
void add(int w, int d) {
  for (; w <= n; g[w] += d, w += w & -w)
    ;
}
int ask(int w) {
  int ans = 0;
  for (; w; ans += g[w], w ^= w & -w)
    ;
  return ans;
}
void check(int k, int l, int r) {
  t[k] += mk[k];
  if (l != r) {
    mk[k * 2] += mk[k];
    mk[k * 2 + 1] += mk[k];
  }
  mk[k] = 0;
}
int calc(int id) {
  int ans = lower_bound(b + 1, b + m + 1, h - a[id]) - b;
  return ask(to[id]) - ans;
}
void ins(int k, int l, int r, int x, int y, int key, int type) {
  check(k, l, r);
  if (y < l || r < x) return;
  if (x <= l && r <= y) {
    if (!type)
      t[k] = key;
    else {
      mk[k] += key;
      check(k, l, r);
    }
    return;
  }
  int mid = l + r >> 1;
  ins(k * 2, l, mid, x, y, key, type);
  ins(k * 2 + 1, mid + 1, r, x, y, key, type);
  t[k] = min(t[k * 2], t[k * 2 + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  fill(t, t + Maxn * 3, oo);
  cin >> n >> m >> h;
  for (int i = 1; i <= m; i++) cin >> b[i];
  for (int i = 1; i <= n; i++) cin >> a[i], c[i].first = a[i], c[i].second = i;
  sort(b + 1, b + m + 1);
  sort(c + 1, c + n + 1, cmp);
  for (int i = 1; i <= n; i++) to[c[i].second] = i;
  for (int i = 1; i <= m; i++) add(to[i], 1);
  for (int i = 1; i <= m; i++) ins(1, 1, n, to[i], to[i], calc(i), 0);
  int ans = t[1] >= 0;
  for (int i = m + 1; i <= n; i++) {
    add(to[i], 1);
    add(to[i - m], -1);
    ins(1, 1, n, to[i - m], to[i - m], oo, 0);
    ins(1, 1, n, to[i - m] + 1, n, -1, 1);
    ins(1, 1, n, to[i] + 1, n, 1, 1);
    ins(1, 1, n, to[i], to[i], calc(i), 0);
    ans += t[1] >= 0;
  }
  cout << ans << endl;
}
