Codeforces Round #441 (Div. 2, by Moscow Team Olympiad) F. High Cry（思维 统计）

F. High Cry

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Disclaimer: there are lots of untranslateable puns in the Russian version of the statement, so there is one more reason for you to learn Russian :)

Rick and Morty like to go to the ridge High Cry for crying loudly — there is an extraordinary echo. Recently they discovered an interesting acoustic characteristic of this ridge: if Rick and Morty begin crying simultaneously from different mountains, their cry would be heard between these mountains up to the height equal the bitwise OR of mountains they've climbed and all the mountains between them.

Bitwise OR is a binary operation which is determined the following way. Consider representation of numbers x and y in binary numeric system (probably with leading zeroes) x = xk... x1x0 and y = yk... y1y0. Then z = x | y is defined following way: z = zk... z1z0, where zi = 1, if xi = 1 or yi = 1, and zi = 0 otherwise. In the other words, digit of bitwise OR of two numbers equals zero if and only if digits at corresponding positions is both numbers equals zero. For example bitwise OR of numbers 10 = 10102 and 9 = 10012 equals 11 = 10112. In programming languages C/C++/Java/Python this operation is defined as «|», and in Pascal as «or».

Help Rick and Morty calculate the number of ways they can select two mountains in such a way that if they start crying from these mountains their cry will be heard above these mountains and all mountains between them. More formally you should find number of pairs land r (1 ≤ l < r ≤ n) such that bitwise OR of heights of all mountains between l and r (inclusive) is larger than the height of any mountain at this interval.

Input

The first line contains integer n (1 ≤ n ≤ 200 000), the number of mountains in the ridge.

Second line contains n integers ai (0 ≤ ai ≤ 109), the heights of mountains in order they are located in the ridge.

Output

Print the only integer, the number of ways to choose two different mountains.

Examples

input

5
3 2 1 6 5

output

8

input

4
3 3 3 3

output

0

Note

In the first test case all the ways are pairs of mountains with the numbers (numbering from one):

(1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)

In the second test case there are no such pairs because for any pair of mountains the height of cry from them is 3, and this height is equal to the height of any mountain.

【题意】给你一个序列，问有多少个区间满足区间所有数的或值大于区间的所有数。

【分析】或值大于区间的所有数，即大于这个区间内的最大值。那我们可以枚举每个数作为最大值来算他的贡献。那么对于每个数，我们就要找到一个极限的区间使得这个数在此区间内是最大值。那么如何知道某个数在这个区间内或值大于他呢？考虑到或的性质，如果这个数 x 第 i 位上为0，而，离他最近的某个数 y 第 i 位上为1，那么区间内包含这个数x就必须包含这个数y,那么对于每一个非1 位，取一个离他最近的，左边右边都是。以上就是两个预处理，第一个预处理求每个数找到左边第一个大于等于它的数的位置的后一位置，和右边第一个大于它的数的前一位置。第二个预处理求满足或条件的最小区间，然后算贡献即可。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 2e5+;;
const int M = ;
const int mod = ;
const int mo=;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
typedef pair<ll,int>P;
int n,m,k;
int a[N],l[N],r[N],L[N],R[N],st[N],pos[];
int main(){
    scanf("%d",&n);
    for(int i=;i<=n;i++){
        scanf("%d",&a[i]);
        l[i]=;r[i]=n;
        L[i]=;R[i]=n+;
    }
    for(int i=;i<=n;i++){
        while(st[]>&&a[st[st[]]]<a[i])--st[];
        if(st[])l[i]=st[st[]]+;
        st[++st[]]=i;
    }
    st[]=;
    for(int i=n;i>;i--){
        while(st[]>&&a[st[st[]]]<=a[i])--st[];
        if(st[])r[i]=st[st[]]-;
        st[++st[]]=i;
    }
    for(int i=;i<=n;i++){
        for(int j=;j<=;j++){
            if((a[i]&(<<j))==){
                L[i]=max(L[i],pos[j]);
            }
            else {
                pos[j]=i;
            }
        }
    }
    for(int i=;i<=;i++)pos[i]=n+;
    for(int i=n;i>;i--){
        for(int j=;j<=;j++){
            if((a[i]&(<<j))==){
                R[i]=min(R[i],pos[j]);
            }
            else {
                pos[j]=i;
            }
        }
    }
    ll ans=;
    for(int i=;i<=n;i++){
        if(l[i]<=L[i])ans+=1LL*(L[i]-l[i]+)*(r[i]-i+);
        if(r[i]>=R[i])ans+=1LL*(r[i]-R[i]+)*(i-l[i]+);
        if(l[i]<=L[i]&&r[i]>=R[i])ans-=1LL*(L[i]-l[i]+)*(r[i]-R[i]+);
    }
    printf("%lld\n",ans);
    return ;
}
/*
5
3 2 1 6 5
*/