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 N = 150010;
int m, n, h;
int a[N], b[N];
struct node {
  int l, r, lazy, v;
} tree[N << 2];
inline void pushup(int x) {
  tree[x].v = min(tree[x << 1].v, tree[x << 1 | 1].v);
}
void build(int l, int r, int x) {
  tree[x].l = l;
  tree[x].r = r;
  if (l == r) {
    tree[x].v = n - l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, x << 1);
  build(mid + 1, r, x << 1 | 1);
  pushup(x);
}
inline void pushdown(int x) {
  if (tree[x].lazy && tree[x].l != tree[x].r) {
    tree[x << 1].v += tree[x].lazy;
    tree[x << 1].lazy += tree[x].lazy;
    tree[x << 1 | 1].v += tree[x].lazy;
    tree[x << 1 | 1].lazy += tree[x].lazy;
  }
  tree[x].lazy = 0;
}
void updata(int l, int r, int v, int x) {
  if (l == tree[x].l && tree[x].r == r) {
    tree[x].lazy += v;
    tree[x].v += v;
    return;
  }
  pushdown(x);
  int ls = x << 1, rs = ls | 1, mid = (tree[x].l + tree[x].r) >> 1;
  if (l > mid)
    updata(l, r, v, rs);
  else if (r <= mid)
    updata(l, r, v, ls);
  else
    updata(l, mid, v, ls), updata(mid + 1, r, v, rs);
  pushup(x);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> m >> n >> h;
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= m; i++) cin >> a[i];
  build(1, n, 1);
  sort(b + 1, b + 1 + n);
  int ans = 0;
  for (int i = 1; i <= m; i++) {
    int l = lower_bound(b + 1, b + 1 + n, h - a[i]) - b;
    if (l <= n) updata(l, n, 1, 1);
    if (i > n) {
      l = lower_bound(b + 1, b + 1 + n, h - a[i - n]) - b;
      if (l <= n) updata(l, n, -1, 1);
    }
    if (tree[1].v >= n) ++ans;
  }
  cout << ans << endl;
  return 0;
}
