Problem:
Score : 300 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

Find the sum of A_i \times A_j over all pairs (i,j) such that 1\leq i < j \leq N, modulo (10^9+7).

Constraints

2 \leq N \leq 2\times 10^5

0 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

Print \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} A_i A_j, modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

11

We have 1 \times 2 + 1 \times 3 + 2 \times 3 = 11.

Sample Input 2

4
141421356 17320508 22360679 244949

Sample Output 2

437235829

Solution in OCaml (4.10.0) (judge verdict: Accepted)

let as_num_list s =
  let splitted = String.split_on_char ' ' s in
  List.map int_of_string splitted

(* 累積和を作る *)
let rec make_cumulative_sum sum_a a_list = match a_list with
    [] -> []
  | first :: rest ->
     let sumMinusFirst = sum_a - first in
     let nextSum = if sumMinusFirst < 0
                   then sumMinusFirst + 1000000007
                   else sumMinusFirst in
     nextSum :: make_cumulative_sum nextSum rest
  
let find_sum_of_product_of_pairs a_list cumulative_sum =
  List.fold_left2
    (fun acc x y -> (acc + x * y) mod 1000000007)
    0
    a_list
    cumulative_sum

let main () =
  let _ = Scanf.sscanf (read_line ()) "%d" (fun x -> x) in (* nは飛ばす *)
  let as_str = read_line () in
  let a_list = as_num_list as_str in
  let sum_a = List.fold_left (fun acc x -> (x + acc) mod 1000000007) 0 a_list in
  let cumulative_sum = make_cumulative_sum sum_a a_list in
  let ans = find_sum_of_product_of_pairs a_list cumulative_sum in
  (
    Printf.printf "%d" ans;
    print_newline ()
  )

let _ = main ()