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;
template <typename T>
inline T qmin(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T qmax(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline void getmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <typename T>
inline void getmax(T &a, const T &b) {
  if (a < b) a = b;
}
inline void fileio(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const int inf = (int)1e9 + 7;
const long long linf = (long long)1e17 + 7;
const int N = 2e5 + 7, Node = N << 2;
int mx[Node], dt[Node];
int a[N], b[N], n, m, h;
inline void settag(int u, int v) { mx[u] += v, dt[u] += v; }
inline void upd(int u) { mx[u] = qmax(mx[((u) << 1)], mx[((u) << 1 ^ 1)]); }
inline void psd(int u) {
  if (!dt[u]) return;
  settag(((u) << 1), dt[u]);
  settag(((u) << 1 ^ 1), dt[u]);
  dt[u] = 0;
}
inline void init(int u, int l, int r) {
  if (l == r) {
    mx[u] = m - l + 1;
    return;
  }
  int mid = l + r >> 1;
  init(((u) << 1), l, mid);
  init(((u) << 1 ^ 1), mid + 1, r);
  upd(u);
}
inline void ins(int u, int l, int r, int l1, int r1, int v) {
  if (l1 > r1) return;
  if (l1 <= l && r <= r1) {
    settag(u, v);
    return;
  }
  psd(u);
  int mid = l + r >> 1;
  if (l1 <= mid) ins(((u) << 1), l, mid, l1, r1, v);
  if (r1 > mid) ins(((u) << 1 ^ 1), mid + 1, r, l1, r1, v);
  upd(u);
}
inline int pos(int v) { return upper_bound(b + 1, b + m + 1, v) - b - 1; }
int main() {
  scanf("%d%d%d", &n, &m, &h);
  for (register int i = (1); i <= (m); ++i) scanf("%d", b + i), b[i] = h - b[i];
  for (register int i = (1); i <= (n); ++i) scanf("%d", a + i);
  sort(b + 1, b + m + 1);
  init(1, 1, m);
  for (register int i = (1); i <= (m - 1); ++i) ins(1, 1, m, 1, pos(a[i]), -1);
  int ans = 0;
  for (register int i = (m); i <= (n); ++i) {
    ins(1, 1, m, 1, pos(a[i]), -1);
    ans += (mx[1] <= 0);
    ins(1, 1, m, 1, pos(a[i - m + 1]), 1);
  }
  printf("%d\n", ans);
  return 0;
}
