// Author: btjanaka (Bryon Tjanaka)

// Problem: (UVa) 10037

#include <bits/stdc++.h>

#define FOR(i, a, b) for (int i = a; i < b; ++i)

#define FORe(i, a, b) for (int i = a; i <= b; ++i)

#define PAI(arr, len) /*Print array of integers*/ \

{ \

for (int _i = 0; _i < len; ++_i) { \

if (_i != len - 1) { \

printf("%d ", arr[_i]); \

} else { \

printf("%d", arr[_i]); \

} \

} \

putchar('\n'); \

}

#define PBS(n, len) /*Print a bitset*/ \

{ \

for (int _i = 0; _i < len; ++_i) { \

putchar(n % 2 + '0'); \

n /= 2; \

} \

putchar('\n'); \

}

#define GET(x) scanf("%d", &x)

#define PLN putchar('\n')

#define INF 2147483647

typedef long long ll;

using namespace std;

int ppl[1010];

pair<int, int> data[10000];

int main() {

int ca;

GET(ca);

while (ca--) {

int n;

GET(n);

FOR(i, 0, n) { GET(ppl[i]); }

sort(ppl, ppl + n);

// See

// https://codingstrife.wordpress.com/2013/07/23/solution-uva10037-pc110403-bridge/

// for algorithm

int tot = 0;

int d_i = 0;

// Move all the slowest members over

// In each case, ppl[0] and ppl[1] start on the uncrossed side. ppl[i] and

// ppl[i + 1] are the slow members that need to cross.

// Method 1: fastest person does all the moving

// Method 2: fastest and next fastest work together

int limit = (n % 2 == 0) ? 2 : 3;

for (int i = n - 2; i >= limit; i -= 2) {

int method1 = ppl[i] + ppl[0] + ppl[i + 1] + ppl[0];

int method2 = ppl[1] + ppl[0] + ppl[i + 1] + ppl[1];

if (method1 < method2) {

tot += method1;

data[d_i++] = make_pair(ppl[0], ppl[i]);

data[d_i++] = make_pair(ppl[0], -1);

data[d_i++] = make_pair(ppl[0], ppl[i + 1]);

data[d_i++] = make_pair(ppl[0], -1);

} else {

tot += method2;

data[d_i++] = make_pair(ppl[0], ppl[1]);

data[d_i++] = make_pair(ppl[0], -1);

data[d_i++] = make_pair(ppl[i], ppl[i + 1]);

data[d_i++] = make_pair(ppl[1], -1);

}

n -= 2;

}

// There should now be 1, 2, or 3 members left, and they should all be the

// fastest of the original group. There is only one optimal crossing for

// each of these cases.

switch (n) {

case 1:

data[d_i++] = make_pair(ppl[0], -1);

tot += ppl[0];

break;

case 2:

data[d_i++] = make_pair(ppl[0], ppl[1]);

tot += ppl[1];

break;

case 3:

data[d_i++] = make_pair(ppl[0], ppl[1]);

data[d_i++] = make_pair(ppl[0], -1);

data[d_i++] = make_pair(ppl[0], ppl[2]);

tot += ppl[2] + ppl[0] + ppl[1];

break;

}

// Output

printf("%d\n", tot);

FOR(i, 0, d_i) {

printf("%d", data[i].first);

if (data[i].second != -1) printf(" %d", data[i].second);

PLN;

}

if (ca) PLN;

}

return 0;

}