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;
#pragma GCC optimize("Ofast")
const long long maxn = 2e5 + 10;
const long long mod = 998244353;
const long long inf = 1e9 + 10;
long long n, m, q;
long long mn[maxn << 2], seg[maxn << 2], la[maxn << 2];
void shift(long long v) {
  mn[v] += la[v];
  seg[v] += la[v];
  if (v < (maxn << 1)) {
    la[v << 1] += la[v];
    la[v << 1 | 1] += la[v];
  }
  la[v] = 0;
}
void up(long long l, long long r, long long val, long long v = 1,
        long long tl = 1, long long tr = n) {
  shift(v);
  if (l > r) return;
  if (tl == l && tr == r) {
    la[v] += val;
    shift(v);
    return;
  }
  long long tm = (tl + tr) >> 1;
  up(l, min(tm, r), val, v << 1, tl, tm);
  up(max(tm + 1, l), r, val, v << 1 | 1, tm + 1, tr);
  mn[v] = min(mn[v << 1], mn[v << 1 | 1]);
  seg[v] = min(seg[v << 1], seg[v << 1 | 1]);
}
void made(long long pos, char tp, long long v = 1, long long tl = 1,
          long long tr = n) {
  shift(v);
  if (tl == tr) {
    if (tp == 'a')
      mn[v] = seg[v];
    else
      mn[v] = inf;
    return;
  }
  long long tm = (tl + tr) >> 1;
  if (pos <= tm)
    made(pos, tp, v << 1, tl, tm);
  else
    made(pos, tp, v << 1 | 1, tm + 1, tr);
  mn[v] = min(mn[v << 1], mn[v << 1 | 1]);
}
long long a[maxn], b[maxn], c[maxn], pos[maxn], h;
long long qu() {
  shift(1);
  return mn[1];
}
void add(long long i) {
  up(pos[i], n, -1);
  up(pos[i], pos[i], c[i]);
  made(pos[i], 'a');
}
void del(long long i) {
  up(pos[i], n, 1);
  up(pos[i], pos[i], -c[i]);
  made(pos[i], 'r');
}
signed main() {
  cin >> n >> m >> h;
  for (long long i = 1; i <= m; i++) cin >> b[i];
  sort(b + 1, b + m + 1);
  vector<pair<long long, long long>> ve;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    long long x = lower_bound(b + 1, b + m + 1, h - a[i]) - b;
    c[i] = m + 1 - x;
    ve.push_back({c[i], i});
  }
  sort(ve.begin(), ve.end());
  long long I = 0;
  for (auto x : ve) pos[x.second] = ++I;
  memset(mn, 63, sizeof mn);
  long long ans = 0;
  for (long long i = 1; i <= m; i++) add(i);
  if (qu() >= 0) ans++;
  for (long long i = m + 1; i <= n; i++) {
    add(i);
    del(i - m);
    if (qu() >= 0) {
      ans++;
    }
  }
  cout << ans;
}
