Problem:
Score : 500 points

Problem Statement

For two sequences S and T of length N consisting of 0 and 1, let us define f(S, T) as follows:

Consider repeating the following operation on S so that S will be equal to T. f(S, T) is the minimum possible total cost of those operations.

Change S_i (from 0 to 1 or vice versa). The cost of this operation is D \times C_i, where D is the number of integers j such that S_j \neq T_j (1 \leq j \leq N) just before this change.

There are 2^N \times (2^N - 1) pairs (S, T) of different sequences of length N consisting of 0 and 1. Compute the sum of f(S, T) over all of those pairs, modulo (10^9+7).

Constraints

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

1 \leq C_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 C_2 \cdots C_N

Output

Print the sum of f(S, T), modulo (10^9+7).

Sample Input 1

1
1000000000

Sample Output 1

999999993

There are two pairs (S, T) of different sequences of length 2 consisting of 0 and 1, as follows:

S = (0), T = (1): by changing S_1 to 1, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

S = (1), T = (0): by changing S_1 to 0, we can have S = T at the cost of 1000000000, so f(S, T) = 1000000000.

The sum of these is 2000000000, and we should print it modulo (10^9+7), that is, 999999993.

Sample Input 2

2
5 8

Sample Output 2

124

There are 12 pairs (S, T) of different sequences of length 3 consisting of 0 and 1, which include:

S = (0, 1), T = (1, 0)

In this case, if we first change S_1 to 1 then change S_2 to 0, the total cost is 5 \times 2 + 8 = 18. We cannot have S = T at a smaller cost, so f(S, T) = 18.

Sample Input 3

5
52 67 72 25 79

Sample Output 3

269312

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <float.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_MOD			1000000007								// 除数(10の9乗+7)
#define D_ARRAY_MAX		200000									// 最大配列数

// 内部変数
static FILE *szpFpI;											// 入力
static int si1Array[D_ARRAY_MAX];								// 配列
static int siACnt;												// 配列数

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpA;
	static int siTNo;
#endif

// １行出力
int
fOutLine(
	char *pcpLine				// <I> １行
)
{
	char lc1Buf[1024];

#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, pcpLine)) {
		siRes = -1;
	}
#else
	printf("%s", pcpLine);
#endif

	return 0;
}

// ソート関数 - int降順
int
fSortFncID(
	const void *pzpVal1			// <I> 値１
	, const void *pzpVal2		// <I> 値２
)
{
	int *lipVal1 = (int *)pzpVal1;
	int *lipVal2 = (int *)pzpVal2;

	// int降順
	if (*lipVal1 > *lipVal2) {
		return -1;
	}
	else if (*lipVal1 < *lipVal2) {
		return 1;
	}

	return 0;
}

// べき乗 - 取得
int
fGetModPower(
	int piBase					// <I> 基数
	, int piIdx					// <I> 指数
)
{
	// 平方リスト - 作成
	int li1Val[100];
	li1Val[0] = piBase;
	int liCnt = 1;
	int liIdx = 1;
	while (piIdx > liIdx) {
		li1Val[liCnt] = (int)((long long)li1Val[liCnt - 1] * (long long)li1Val[liCnt - 1] % D_MOD);
		liCnt++;
		liIdx += liIdx;
	}

	// べき乗 - 取得
	int liVal = 1;
	while (piIdx > 0) {
		if (piIdx >= liIdx) {
			piIdx -= liIdx;
			liVal = (int)((long long)liVal * (long long)li1Val[liCnt - 1] % D_MOD);
		}
		liCnt--;
		liIdx /= 2;
	}

	return liVal;
}

// 実行メイン
int
fMain(
)
{
	int i;
	char lc1Buf[1024];

	// 配列数 - 取得
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	sscanf(lc1Buf, "%d", &siACnt);

	// 配列 - 取得
	for (i = 0; i < siACnt; i++) {
		fscanf(szpFpI, "%d", &si1Array[i]);
	}
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	qsort(si1Array, siACnt, sizeof(int), fSortFncID);

	// パターン数(１項目) - 取得
	long long llVal = fGetModPower(2, siACnt - 1);
	long long llPCnt1 = llVal * llVal % D_MOD;
	long long llPCnt2 = llPCnt1 * 2 % D_MOD;

	// コスト合計 - 取得
	long long llPSum = llPCnt2;
	long long llCSum = 0;
	for (i = 0; i < siACnt; i++) {
		llCSum = (llCSum + (long long)si1Array[i] * llPSum) % D_MOD;
		llPSum = (llPSum + llPCnt1) % D_MOD;
	}

	return (int)(llCSum % D_MOD);
}

// １回実行
int
fOne(
)
{
	int liRet;
	char lc1Buf[1024];

	// 入力 - セット
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", siTNo);
	szpFpI = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", siTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#else
	szpFpI = stdin;
#endif

	// 実行メイン
	liRet = fMain();

	// 出力
	sprintf(lc1Buf, "%d\n", liRet);
	fOutLine(lc1Buf);

	// 残データ有無
#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, "")) {
		siRes = -1;
	}
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpI);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", siTNo);
	}
	else {
		printf("NG %d\n", siTNo);
	}
#endif

	return 0;
}

// プログラム開始
int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		siTNo = i;
		fOne();
	}
#else
	fOne();
#endif

	return 0;
}