Problem Description

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

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)?=?al | al?+?1 | ... ?| ar.

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

The 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, ..., an (0 ≤ ai ≤ 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

題意：給出 n 個數，定義函數 F(l,r)，表示區間 [l,r] 各項的 或 的和，問 F(l,r) 有多少個不同的值

思路：使用 set，兩重循環枚舉，然后由于所有位都為 1 的數與其他數運算是這個數自身，剪枝剪掉即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+9; const int N = 1000000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;int a[N]; set<int> st; int main() {int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);for(int i=1;i<=n;i++){int x=a[i];int y=0;st.insert(x);for(int j=i+1;j<=n;j++){x|=a[j];y|=a[j];st.insert(x);if(x==y)//剪枝break;}}printf("%d\n",st.size());return 0; }

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