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++14 (GCC 5.4.1) (judge verdict: Accepted)

//Konrad Paluszek,University of Warsaw(former XIV LO Staszic)
//it's nice to stick a fork into an open socket
//clock says it's time to sleep
//our getppid() in heaven, hallowed by ttyname
#ifndef LOCAL
#pragma GCC optimize("O3")
#endif
#define TIME (chrono::steady_clock::now().time_since_epoch().count())
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define xfm(a,b)a##b
#define xwb(a,b)xfm(a,b)
#define _ xwb(nvj,__LINE__)
#define FOR(i,b,e)for(urs((b)+(e))i=(b);i<=(e);++i)
#define REP(i,n)for(urs(n)i=0;i<(n);++i)
#define FORD(i,b,e)for(urs((b)+(e))i=(b);i>=(int)(e);--i)
#define UNTIL(t)while(clock()<(t)*CLOCKS_PER_SEC)
#define PER(r...)for(bool _=1;_||next_permutation(r);_=false)
#define ALL(r)(r).begin(),(r).end()
#define FS(r)r.first,r.second
#define sim template<class c
#define ros return o
#define rans return ans
#define forbits(i,m)if(m)for(urs(m)i=ctz(m),i##nvj=m;i##nvj;i##nvj^=((urs(m))1<<i),i=ctz(i##nvj))
#define fordbits(i,m)if(m)for(urs(m)i=8*sizeof(m)-clz(m)-1,i##nxd=m;i##nxd;i##nxd^=((urs(m))1<<i),i=8*sizeof(m)-clz(i##nxd)-1)
#define ksets(t, m, k, n) for (t m = (((t)1 << (k)) - 1); m < ((t)1 << (n)); m = nux(m))
#define urs(r...)typename decay<decltype(r)>::type
#define hur(f,g,r)sim>int f(c a){if(sizeof(c)==16)return r;if(sizeof(c)==8)return g##ll(a);return g(a);}
#define pwq(t,i) int clz(t x){return clz<int>(x)-i;}
using namespace __gnu_pbds;using namespace std;using ll=long long;using ld=long double;using ull=unsigned long long;using pii=pair<int,int>;using pll=pair<ll,ll>;using vpii=vector<pii>;using unt=unsigned int;sim>using min_queue=priority_queue<c,vector<c>,greater<c>>;sim>using ordered_set=tree<c,null_type,less<c>,rb_tree_tag,tree_order_statistics_node_update>;hur(popc,__builtin_popcount,popc<ull>(a)+popc<ull>(a>>64))hur(ctz,__builtin_ctz,(ull)a?ctz<ull>(a):64+ctz<ull>(a>>64))hur(clz,__builtin_clz,a>>64?clz<ull>(a>>64):64+clz<ull>(a))pwq(short,16)pwq(uint16_t,16)pwq(char,24)pwq(int8_t,24)pwq(uint8_t,24)sim,class N>bool mini(c&o,const N&h){if(o>h)ros=h,1;return 0;}sim,class N>bool maxi(c&o,const N&h){if(o<h)ros=h,1;return 0;}sim,class n>using gyv=c;
#ifdef LOCAL
#include </home/kjp/headers/debuglib.hpp>
#else
#define loc(...)
#define onl(r...)r
#define debug(...)
#define print_stack(...)
#define mark_stack(...)
#define set_pre(...)
#define reg_it(...)
#endif
#define next nexT
#define prev preV
#define tree trEE
#define left lefT
#define right righT
#define div diV
#define y1 y_1
ull mix(ull o){o+=0x9e3779b97f4a7c15;o=(o^(o>>30))*0xbf58476d1ce4e5b9;o=(o^(o>>27))*0x94d049bb133111eb;ros^(o>>31);}ull SALT=0x7a14a4b0881ebf9,tqu=0x7a14a4b0881ebf9;ull my_rand(){return tqu=mix(tqu);}void my_srand(ull x){SALT=tqu=x;}const int inf = 1023400000;const ll llinf=1234567890000000000ll;
#define rand my_rand
#define srand my_srand
#define random_shuffle(r...)random_shuffle(r,[](int _){return my_rand()%_;})
sim>inline c nux(c m){if(!m)return numeric_limits<c>::max();c A=m&-m;c B=~((A-1)^m);c C=B&-B;c D=(C>>(1+ctz(A)))-1;return C|(m&~(C-1))|D;}__attribute__((no_sanitize_undefined))ll mul(ll a,ll b,ll m){ll q=a*(ld)b/m;ll o=a*b-q*m;o%=m;if(o<0)o+=m;ros;}sim>void unq(c&x){x.resize(unique(ALL(x))-x.begin());}
#define q31 (8 * sizeof(c) - 1)
sim> typename enable_if<is_signed<c>::value, c>::type FastMax(c x, c y) { return ((x - y) >> q31 & (x ^ y)) ^ x; }
sim> typename enable_if<is_signed<c>::value, c>::type FastMin(c x, c y) { return ((x - y) >> q31 & (x ^ y)) ^ y; }
sim> typename enable_if<is_signed<c>::value, c>::type FastAbs(c x)        { return (x ^ x >> q31) - (x >> q31); }
sim> typename enable_if<is_signed<c>::value, c>::type gcd(c a, c b) {
	loc(assert(a > 0 && b > 0));
	int aa = ctz(a), bb = ctz(b);
	int r = FastMin(aa, bb);
	a >>= aa;
	while (a && b) {
		b >>= ctz(b);
		c x = FastAbs(a - b);
		a = FastMin(a, b);
		b = x;
	}
	return (a + b) << r;
}
sim> c lcm(c a, c b) {return a / gcd(a, b) * b;}
sim>typename enable_if<is_integral<c>::value,ull>::type my_hash(c x){
	if(sizeof(c)>8) return mix((ull)x^SALT)^mix((ull)(x>>64)^SALT);
	return mix(x^SALT);
}
template<size_t N>ull my_hash(const bitset<N>&x){
	return mix(hash<bitset<N>>()(x)^SALT);
}
sim,class n>ull my_hash(const pair<c,n>&);
sim>gyv<ull,typename c::value_type>my_hash(const c&x){
	ull o=SALT;
	for(auto&d:x)o=mix(o^my_hash(d));
	ros;
}
sim,class n>ull my_hash(const pair<c,n>&x){return mix(my_hash(x.first))^my_hash(x.second);}
sim>struct my_hasher{size_t operator()(const c&o)const{return my_hash(o);}};
sim, class m>
#ifdef LOCAL
using _un0rDer3d_M4P=unordered_map<c,m,my_hasher<c>>;
#else
struct _un0rDer3d_M4P : public cc_hash_table<c,m,my_hasher<c>>{bool count(const c&x)const{return this->find(x) != this->end();}};
#endif
#define unordered_map _un0rDer3d_M4P
#if defined(LOCAL) || defined(LOCAL2)
void __tmi() {cerr << "total time: " << clock() / (ld)CLOCKS_PER_SEC << "s" << endl;}int _ = (atexit(__tmi), 0);
#endif
const int nax = 2e5 + 44, mod = 1e9 + 7;
int a[nax];
ll powe(ll b, ll e) {
	ll r = 1;
	while (e) {
		if (e & 1) r = r * b % mod;
		b = b * b % mod;
		e >>= 1;
	}
	return r;
}
int main() {
	int n;
	scanf("%d", &n);
	REP(i, n) scanf("%d", a + i);
	sort(a, a + n);
	ll ans = 0;
	REP(i, n) {
		ll coef = n + 1 - i;
		ans = (ans + a[i] * coef) % mod;
	}
	ans = ans * powe(2, n * 2 - 2) % mod;
	if (ans < 0) ans += mod;
	printf("%lld\n", ans);
}