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 long long md = 1e9 + 7;
const int xn = 2e5 + 10;
const int xm = 2e1 + 10;
const int SQ = 320;
const int sq = 320 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
long long power(long long a, long long b) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2) % md
                      : power(a * a % md, b / 2) % md));
}
long long zarb(long long a, long long b) { return (a * b + 10 * md) % md; }
long long jaam(long long a, long long b) { return (a + b + 10 * md) % md; }
int n, len, h, part[xn], num[xn], b[xn], a[xn], Seg[xn * 4], Lazy[xn * 4], ans;
void Build(int id, int L, int R) {
  if (R - L == 1) {
    if (L >= 1 && L <= len)
      Seg[id] = -L;
    else
      Seg[id] = inf;
    return;
  }
  int Mid = (L + R) / 2;
  Build(id * 2, L, Mid);
  Build(id * 2 + 1, Mid, R);
  Seg[id] = min(Seg[id * 2], Seg[id * 2 + 1]);
}
void Shift(int id, int L, int R) {
  if (Lazy[id] == 0) return;
  Seg[id] += Lazy[id];
  if (R - L > 1) {
    Lazy[id * 2] += Lazy[id];
    Lazy[id * 2 + 1] += Lazy[id];
  }
  Lazy[id] = 0;
}
void Update(int id, int L, int R, int Ql, int Qr, int val) {
  Shift(id, L, R);
  if (Qr <= L || R <= Ql) return;
  if (Ql <= L && R <= Qr) {
    Lazy[id] += val;
    Shift(id, L, R);
    return;
  }
  int Mid = (L + R) / 2;
  Update(id * 2, L, Mid, Ql, Qr, val);
  Update(id * 2 + 1, Mid, R, Ql, Qr, val);
  Seg[id] = min(Seg[id * 2], Seg[id * 2 + 1]);
}
int GET(int id, int L, int R, int Ql, int Qr) {
  Shift(id, L, R);
  if (Qr <= L || R <= Ql) return inf;
  if (Ql <= L && R <= Qr) return Seg[id];
  int Mid = (L + R) / 2;
  return min(GET(id * 2, L, Mid, Ql, Qr), GET(id * 2 + 1, Mid, R, Ql, Qr));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> len >> h;
  for (int i = 1; i <= len; ++i) cin >> b[i];
  for (int i = 1; i <= n; ++i) cin >> num[i];
  sort(b + 1, b + len + 1);
  for (int i = 1; i <= n; ++i) {
    int L = 0, R = len + 1, Mid;
    while (R - L > 1) {
      Mid = (L + R) / 2;
      if (num[i] + b[Mid] >= h)
        R = Mid;
      else
        L = Mid;
    }
    a[i] = R;
  }
  Build(1, 0, xn);
  for (int i = 1; i <= len; ++i) Update(1, 0, xn, a[i], len + 1, 1);
  ans += (GET(1, 0, xn, 1, len + 1) >= 0);
  for (int i = len + 1; i <= n; ++i) {
    Update(1, 0, xn, a[i - len], len + 1, -1);
    Update(1, 0, xn, a[i], len + 1, 1);
    ans += (GET(1, 0, xn, 1, len + 1) >= 0);
  }
  cout << ans << '\n';
  return 0;
}
