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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
)

func main() {
	var n int
	var a int
	fmt.Scan(&n)
	sliceA := make([]int, n)
	sliceSum := make([]int, n)

	for i := 0; i < n; i++ {
		fmt.Scan(&a)
		sliceA[i] = a
	}

	// 累積和
	sumsum := 0
	for i := 0; i < n; i++ {
		sumsum += sliceA[i]
		sliceSum[i] = sumsum
	}

	sumAll := 0
	mod := 1000000007
	for i := 0; i < n; i++ {
		sumAll += sliceA[i] * (sliceSum[n-1] - sliceSum[i])
		sumAll %= mod
	}
	fmt.Print(sumAll)

	// sumAll := 0
	// mod := 1000000007
	// for i := 0; i < n; i++ {
	// 	sum := (sliceSum[n-1] - sliceSum[i]) % mod
	// 	sumAll += sliceA[i] * sum
	// 	sumAll %= mod
	// }
	// fmt.Print(sumAll)

	// 計算量多い
	// for i := 0; i < n; i++ {
	// 	sum := 0
	// 	for j := i; j < n; j++ {
	// 		if i == j {
	// 			continue
	// 		}
	// 		sum += sliceA[i] * sliceA[j]
	// 	}
	// 	sumAll += sum
	// }
}

func abc(value int) int {
	flag := 1000000007
	ans := 0
	if value < flag {
		ans = value
	} else {
		ans = value % flag
	}
	return ans
}