Revenge of Segment Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1175    Accepted Submission(s): 403

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.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
using namespace std;
const long long MOD=1e9+7;
#define LL long long
LL a[447005],ans;
int t,n;
int main()
{
      scanf("%d",&t);
      while(t--)
      {
            ans=0;
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
                  scanf("%lld",&a[i]);
            for(int i=1;i<=n;i++)
            {
                  ans=(ans+(a[i]*i)%MOD*(n-i+1))%MOD;
            }
            printf("%lld\n",ans);
      }
      return 0;
}