A

You are given an array a consisting of n integers. A subarray (l, r) from array a is defined as non-empty sequence of consecutive elements a_l, a_l + 1, ..., a_r.

The beauty of a subarray (l, r) is calculated as the bitwise AND for all elements in the subarray:

Beauty(l, r) = a_l & a_l + 1 & a_l + 2 & ... & a_r

Your task is to calculate the summation of the beauty of all subarrays (l, r) (1 ≤ l ≤ r ≤ n):

Input

The first line contains an integer T, where T is the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 10⁵), where n is the size of the array a.

The second line of each test case contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁶), giving the array a.

Output

For each test case, print a single line containing the summation of the beauty of all subarrays in the given array.

Example

Input

2
3
7 11 9
4
11 9 6 11

Output

40
48

Note

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

A bitwise AND takes two equal-length binary representations and performs the logical AND operation on each pair of the corresponding bits, by multiplying them. Thus, if both bits in the compared position are 1, the bit in the resulting binary representation is 1 (1  ×  1 = 1); otherwise, the result is 0 (1  ×  0 = 0 and 0  ×  0 = 0). This operation exists in all modern programming languages, for example in language C++ and Java it is marked as &.

In the first test case, the answer is calculated as summation of 6 subarrays as follow:

Beauty(1, 1) + Beauty(l, 2) + Beauty(1, 3) + Beauty(2, 2) + Beauty(2, 3) + Beauty(3, 3) (7) + (7 & 11) + (7 & 11 & 9) + (11) + (11 & 9) + (9) = 40 这个题目呢，看完题解就觉得不难了，我开始觉得这个比较难，因为对于位运算不是很熟悉，现在明白了，这种位运算最后转化成二进制去求解比较好，这个有个要求就是只对于连续的才去进行位运算，所以我们就把这个每一个数的位运算的值求出来，然后再把连续的进行位运算。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 100;
int num[maxn][30];

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            int x;
            scanf("%d", &x);
            for (int j = 0; j <= 20; j++)
            {
                num[i][j] = (x >> j) & 1;
            }
        }
        int cnt = 0;
        ll ans = 0;
        for (int j = 0; j <= 20; j++)
        {
            cnt = 0;
            for (int i = 1; i <= n; i++)
            {
                if (num[i][j]) cnt++;
                else
                {
                    ans += 1ll * cnt*(cnt + 1) / 2 * (1ll << j);
                    cnt = 0;
                }
            }
            if (cnt) ans += 1ll * cnt*(cnt + 1) / 2 * (1ll << j);
        }
        printf("%I64d
", ans);
    }
    return 0;
}