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;
template <int MAXN>
struct segtree_lazy {
  struct updater {
    int x;
    updater(int x = 0) : x(x) {}
    updater& operator+=(const updater& other) {
      x += other.x;
      return *this;
    }
    operator bool() const { return x != 0; }
  };
  struct node_t {
    int x;
    node_t(int x = 1123123123) : x(x) {}
    node_t& operator+=(const node_t& other) {
      x = min(x, other.x);
      return *this;
    }
    node_t& operator+=(const updater& other) {
      x += other.x;
      return *this;
    }
    node_t operator+(const node_t& other) const {
      node_t tmp = *this;
      tmp += other;
      return tmp;
    }
  };
  node_t a[2 * MAXN];
  updater b[2 * MAXN];
  void init() {
    for (int i = 0; i < 2 * MAXN; i++) a[i] = node_t(0);
  }
  void push(int i) {
    if (b[i]) {
      a[i] += b[i];
      if (i < MAXN) {
        b[2 * i] += b[i];
        b[2 * i + 1] += b[i];
      }
      b[i] = updater();
    }
  }
  node_t get(int l, int r, int node = 1, int nl = 0, int nr = MAXN - 1) {
    push(node);
    if (r < nl || nr < l) {
      return node_t();
    }
    if (l <= nl && nr <= r) {
      return a[node];
    }
    int nm = (nl + nr) >> 1;
    return get(l, r, 2 * node, nl, nm) + get(l, r, 2 * node + 1, nm + 1, nr);
  }
  void update(int l, int r, updater val, int node = 1, int nl = 0,
              int nr = MAXN - 1) {
    push(node);
    if (r < nl || nr < l) {
      return;
    }
    if (l <= nl && nr <= r) {
      b[node] += val;
      push(node);
      return;
    }
    int nm = (nl + nr) >> 1;
    update(l, r, val, 2 * node, nl, nm);
    update(l, r, val, 2 * node + 1, nm + 1, nr);
    a[node] = a[2 * node] + a[2 * node + 1];
  }
};
segtree_lazy<262144> drvo;
int n, m;
int a[150005], b[150005], h;
void dodaj(int x, int z) {
  auto it = lower_bound(b, b + m, h - x) - b;
  drvo.update(it, m - 1, z);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  drvo.init();
  cin >> n >> m >> h;
  for (int i = 0; i < m; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(b, b + m);
  for (int i = 0; i < m; i++) drvo.update(i, m - 1, -1);
  for (int j = 0; j < m; j++) dodaj(a[j], 1);
  int sol = 0;
  for (int i = 0; i <= n - m; i++) {
    sol += drvo.get(0, m - 1).x >= 0;
    dodaj(a[i + m], 1);
    dodaj(a[i], -1);
  }
  cout << sol << '\n';
}
