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.

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

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

2
1
2
3
1 2 3

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。
那么包括Ai的区间满足。累加就能够了。

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;
}