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 MAX_SIZE = 150100;
struct Node {
  int left, right;
  int l, r;
  int val;
  bool lazy;
  int inc;
};
Node tree[MAX_SIZE << 3];
int vn;
int na, nb, lim;
int a[MAX_SIZE], b[MAX_SIZE];
int low[MAX_SIZE];
int root;
inline void update(int, int);
inline void pushDown(int);
inline void update(int cur, int inc) {
  if (tree[cur].lazy) {
    tree[cur].val += inc;
    tree[cur].inc += inc;
  } else {
    tree[cur].val += inc;
    tree[cur].lazy = true;
    tree[cur].inc = inc;
  }
}
inline void pushDown(int cur) {
  if (tree[cur].l == tree[cur].r) return;
  update(tree[cur].left, tree[cur].inc);
  update(tree[cur].right, tree[cur].inc);
  tree[cur].lazy = false;
}
int build(int l, int r) {
  int cur = ++vn;
  tree[cur].l = l;
  tree[cur].r = r;
  tree[cur].lazy = false;
  tree[cur].inc = 0;
  if (l == r) {
    tree[cur].val = -l - 1;
    tree[cur].left = tree[cur].right = 0;
  } else {
    int mid = (l + r) >> 1;
    tree[cur].left = build(l, mid);
    tree[cur].right = build(mid + 1, r);
    tree[cur].val = min(tree[tree[cur].left].val, tree[tree[cur].right].val);
  }
  return cur;
}
void modify(int cur, int l, int r, int inc) {
  if (l > r) return;
  if (tree[cur].lazy) pushDown(cur);
  if (tree[cur].l == l && tree[cur].r == r)
    update(cur, inc);
  else {
    int mid = (tree[cur].l + tree[cur].r) >> 1;
    if (r <= mid)
      modify(tree[cur].left, l, r, inc);
    else if (l > mid)
      modify(tree[cur].right, l, r, inc);
    else {
      modify(tree[cur].left, l, mid, inc);
      modify(tree[cur].right, mid + 1, r, inc);
    }
    tree[cur].val = min(tree[tree[cur].left].val, tree[tree[cur].right].val);
  }
}
int main() {
  scanf("%d%d%d", &na, &nb, &lim);
  vn = 0;
  for (int i = 0; i < nb; i++) scanf("%d", b + i);
  sort(b, b + nb);
  for (int i = 0; i < na; i++) {
    scanf("%d", a + i);
    low[i] = lower_bound(b, b + nb, lim - a[i]) - b;
  }
  root = build(0, nb - 1);
  for (int i = 0; i < nb; i++) modify(root, low[i], nb - 1, 1);
  int ans = 0;
  for (int i = 0; i <= na - nb; i++) {
    ans += (tree[root].val >= 0);
    if (i != na - nb) {
      modify(root, low[i], nb - 1, -1);
      modify(root, low[i + nb], nb - 1, 1);
    }
  }
  printf("%d\n", ans);
  return 0;
}
