CodeForces - 243A——The Brand New Function （令人烦躁的思维题）

Polycarpus has a sequence, consisting of n non-negative integers: a₁, a₂, ..., a_n.

Let's define function f(l, r) (l, r are integer, 1 ≤ l ≤ r ≤ n) for sequence a as an operation of bitwise OR of all the sequence elements with indexes from l to r. Formally: f(l, r) = a_l | a_l + 1 | ...  | a_r.

Polycarpus took a piece of paper and wrote out the values of function f(l, r) for all l, r (l, r are integer, 1 ≤ l ≤ r ≤ n). Now he wants to know, how many distinct values he's got in the end.

Help Polycarpus, count the number of distinct values of function f(l, r) for the given sequence a.

Expression x | y means applying the operation of bitwise OR to numbers x and y. This operation exists in all modern programming languages, for example, in language C++ and Java it is marked as "|", in Pascal — as "or".

Input

he first line contains integer n (1 ≤ n ≤ 10⁵) — the number of elements of sequence a. The second line contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁶) — the elements of sequence a.

Output

Print a single integer — the number of distinct values of function f(l, r) for the given sequence a.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Examples

Input

3
1 2 0

Output

4

Input

10
1 2 3 4 5 6 1 2 9 10

Output

11

Note

In the first test case Polycarpus will have 6 numbers written on the paper: f(1, 1) = 1, f(1, 2) = 3, f(1, 3) = 3, f(2, 2) = 2, f(2, 3) = 2, f(3, 3) = 0. There are exactly 4 distinct numbers among them: 0, 1, 2, 3.

题解：

这个题是真的迷。首先区间DP的话空间不够，枚举估计也要超时。当时真是想了各种方法都不行。之后一看别人的AC代码，WTFC！。两层for循环加一个剪枝就过去了。相当于是个O（NlogN）复杂度。这TM不就是测试数据水吗？？？真的烦。

代码：

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 100005;

int board[MAXN];

set<int> S;

int main(){
	
	int N;
	while(scanf("%d",&N)!=EOF){
		S.clear();
		for(int i=0 ; i<N ; i++){
			scanf("%d",&board[i]);
		}
		for(int i=0 ; i<N ; i++){
			int a = board[i];
			int b = 0;
			S.insert(a);
			for(int j=i+1 ; j<N ; j++){
				a |= board[j]; 
				b |= board[j];
				S.insert(a);
				if(a == b)break;//这里剪枝减掉了当已有数进行位或运算达到所有位都为1后的运算。
				//如果a本身就等于0那么也直接减掉就行，不影响结果。 
			}
		}
		printf("%d\n",S.size());
	}
	
	return 0;
}