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 int inf = 0x7fffffff;
const int N = 150010;
struct s_tree {
  int _min[N << 2], tag[N << 2];
  void pushdown(int p, int L, int R) {
    int l = (((p) << 1) + 1), r = (((p) << 1) + 2);
    _min[p] += tag[p];
    int d = tag[p];
    tag[p] = 0;
    if (L == R) return;
    tag[l] += d;
    tag[r] += d;
  }
  void upd(int p, int L, int R) {
    int l = (((p) << 1) + 1), r = (((p) << 1) + 2);
    pushdown(l, L, L + R >> 1);
    pushdown(r, ((L + R) >> 1) + 1, R);
    _min[p] = min(_min[l], _min[r]);
  }
  void insert(int p, int l, int r, int h, int e, int c) {
    pushdown(p, l, r);
    if (l == h && r == e) {
      tag[p] += c;
      return;
    }
    int mid = l + r >> 1;
    if (h <= mid) insert((((p) << 1) + 1), l, mid, h, min(mid, e), c);
    if (e > mid) insert((((p) << 1) + 2), mid + 1, r, max(mid + 1, h), e, c);
    upd(p, l, r);
  }
  int query(int p, int l, int r, int h, int e) {
    pushdown(p, l, r);
    if (l == h && r == e) {
      return _min[p];
    }
    int al, ar;
    al = ar = inf;
    int mid = l + r >> 1;
    if (h <= mid) al = query((((p) << 1) + 1), l, mid, h, min(mid, e));
    if (e > mid) ar = query((((p) << 1) + 2), mid + 1, r, max(mid + 1, h), e);
    upd(p, l, r);
    return min(al, ar);
  }
} st;
int n, len, h, a[N], b[N], d[N];
int main() {
  scanf("%d%d%d", &n, &len, &h);
  for (int i = 0; i < len; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(b, b + len);
  for (int i = 0; i < n; i++) {
    int l = 0, r = len - 1, mid, t = -1;
    while (l <= r) {
      mid = l + r >> 1;
      if (b[mid] + a[i] >= h)
        t = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    if (t != -1)
      d[i] = t;
    else
      d[i] = len;
  }
  for (int i = 0; i < len; i++) st.insert(0, 0, len, i, i, len - i);
  int ans = 0;
  for (int i = 0; i < len; i++) {
    st.insert(0, 0, len, 0, d[i], -1);
  }
  if (st.query(0, 0, len, 0, len) >= 0) ans++;
  for (int i = 1; i + len - 1 < n; i++) {
    st.insert(0, 0, len, 0, d[i - 1], 1);
    st.insert(0, 0, len, 0, d[i + len - 1], -1);
    if (st.query(0, 0, len, 0, len) >= 0) {
      ans++;
    }
  }
  printf("%d\n", ans);
}
