Problem Description

In computer science, a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point. It is, in principle, a static structure; that is, its content cannot be modified once the
structure is built. A similar data structure is the interval tree.

A segment tree for a set I of n intervals uses O(n log n) storage and can be built in O(n log n) time. Segment trees support searching for all the intervals that contain a query point in O(log n + k), k being the number of retrieved intervals or segments.

---Wikipedia



Today, Segment Tree takes revenge on you. As Segment Tree can answer the sum query of a interval sequence easily, your task is calculating the sum of the sum of all continuous sub-sequences of a given number sequence.

 

Input

The first line contains a single integer T, indicating the number of test cases. 



Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.



[Technical Specification]

1. 1 <= T <= 10

2. 1 <= N <= 447 000

3. 0 <= Ai <= 1 000 000 000

 

Output

For each test case, output the answer mod 1 000 000 007.

 

Sample Input

2

1

2

3

1 2 3

 

Sample Output

2

20

Hint

For the second test case, all continuous sub-sequences are [1], [2], [3], [1, 2], [2, 3] and [1, 2, 3]. So the sum of the sum of the sub-sequences is 1 + 2 + 3 + 3 + 5 + 6 = 20.

Huge input, faster I/O method is recommended. And as N is rather big, too straightforward algorithm (for example, O(N^2)) will lead Time Limit Exceeded.

And one more little helpful hint, be careful about the overflow of int.

 

考虑每一个数出如今多少个子序列之中。如果第i个数为Ai。区间为[L,R]。

那么包括Ai的区间满足L⩽i⋂R⩾i。累加(L+1)∗(N−R)∗A[i]就能够了。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

long long mod=1000000000+7;

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        long long n;

        long long ans=0;

        long long a;

        scanf("%I64d",&n);

        for(int i=0;i<n;i++)

        {

            scanf("%I64d",&a);

            ans=(ans+(((a*(i+1)%mod)*(n-i))%mod))%mod;

        }

        printf("%I64d\n",ans);

    }

    return 0;

}