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;
const long long N = 2e5 + 5;
const long long inf = 1e16 + 7;
long long m, n, h;
long long ans = 0;
long long seg[4 * N];
long long lazy[4 * N];
long long ind[N];
vector<long long> b;
vector<long long> a;
void build(long long p, long long val, long long v = 1, long long s = 0,
           long long e = n + 1) {
  if (e - s == 1) {
    seg[v] = val;
    return;
  }
  long long mid = (e + s) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  if (p < mid) {
    build(p, val, lc, s, mid);
  } else {
    build(p, val, rc, mid, e);
  }
  seg[v] = min(seg[lc], seg[rc]);
}
void shift(long long v, long long s, long long e) {
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  seg[lc] += lazy[v];
  seg[rc] += lazy[v];
  lazy[lc] += lazy[v];
  lazy[rc] += lazy[v];
  lazy[v] = 0;
}
void add(long long l, long long r, long long val, long long v = 1,
         long long s = 0, long long e = n + 1) {
  if (r <= s || e <= l) {
    return;
  }
  if (l <= s && e <= r) {
    seg[v] += val;
    lazy[v] += val;
    return;
  }
  shift(v, s, e);
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  add(l, r, val, lc, s, mid);
  add(l, r, val, rc, mid, e);
  seg[v] = min(seg[lc], seg[rc]);
}
long long get(long long l, long long r, long long v = 1, long long s = 0,
              long long e = n + 1) {
  if (r <= s || e <= l) {
    return inf;
  }
  if (l <= s && e <= r) {
    return seg[v];
  }
  shift(v, s, e);
  long long mid = (s + e) / 2;
  long long lc = 2 * v;
  long long rc = lc + 1;
  return min(get(l, r, lc, s, mid), get(l, r, rc, mid, e));
}
int main() {
  memset(seg, 127, sizeof seg);
  cin >> m >> n >> h;
  for (long long i = 0; i < n; i++) {
    long long tmp;
    cin >> tmp;
    b.push_back(tmp);
  }
  for (long long i = 0; i < m; i++) {
    long long tmp;
    cin >> tmp;
    a.push_back(tmp);
  }
  sort(b.begin(), b.end());
  for (long long i = 0; i < m; i++) {
    ind[i] = lower_bound(b.begin(), b.end(), h - a[i]) - b.begin();
  }
  for (long long i = 0; i <= n; i++) {
    build(i, n - i);
  }
  for (long long i = 0; i <= n; i++) {
  }
  for (long long i = 0; i < n; i++) {
    add(0, ind[i] + 1, -1);
  }
  if (get(0, n + 1) >= 0) {
    ans++;
  }
  for (long long i = n; i < m; i++) {
    add(0, ind[i - n] + 1, +1);
    add(0, ind[i] + 1, -1);
    if (get(0, n + 1) >= 0) {
      ans++;
    }
  }
  cout << ans << endl;
}
