Hanh lives in a shared apartment. There are n people (including Hanh) living there, each has a private fridge. 

n fridges are secured by several steel chains. Each steel chain connects two different fridges and is protected by a digital lock. The owner of a fridge knows passcodes of all chains connected to it. A fridge can be open only if all chains connected to it are unlocked. For example, if a fridge has no chains connected to it at all, then any of n people can open it.

<image> For exampe, in the picture there are n=4 people and 5 chains. The first person knows passcodes of two chains: 1-4 and 1-2. The fridge 1 can be open by its owner (the person 1), also two people 2 and 4 (acting together) can open it.

The weights of these fridges are a_1, a_2, …, a_n. To make a steel chain connecting fridges u and v, you have to pay a_u + a_v dollars. Note that the landlord allows you to create multiple chains connecting the same pair of fridges. 

Hanh's apartment landlord asks you to create exactly m steel chains so that all fridges are private. A fridge is private if and only if, among n people living in the apartment, only the owner can open it (i.e. no other person acting alone can do it). In other words, the fridge i is not private if there exists the person j (i ≠ j) that the person j can open the fridge i.

For example, in the picture all the fridges are private. On the other hand, if there are n=2 fridges and only one chain (which connects them) then both fridges are not private (both fridges can be open not only by its owner but also by another person).

Of course, the landlord wants to minimize the total cost of all steel chains to fulfill his request. Determine whether there exists any way to make exactly m chains, and if yes, output any solution that minimizes the total cost. 

Input

Each test contains multiple test cases. The first line contains the number of test cases T (1 ≤ T ≤ 10). Then the descriptions of the test cases follow.

The first line of each test case contains two integers n, m (2 ≤ n ≤ 1000, 1 ≤ m ≤ n) — the number of people living in Hanh's apartment and the number of steel chains that the landlord requires, respectively.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^4) — weights of all fridges.

Output

For each test case:

  * If there is no solution, print a single integer -1. 
  * Otherwise, print a single integer c — the minimum total cost. The i-th of the next m lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i), meaning that the i-th steel chain connects fridges u_i and v_i. An arbitrary number of chains can be between a pair of fridges. 



If there are multiple answers, print any.

Example

Input


3
4 4
1 1 1 1
3 1
1 2 3
3 3
1 2 3


Output


8
1 2
4 3
3 2
4 1
-1
12
3 2
1 2
3 1
"""
    Satwik_Tiwari ;) .
    12th Sept , 2020  - Saturday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase

from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl

#If the element is already present in the list,
# the left most position where element has to be inserted is returned.

from bisect import bisect_right as br
from bisect import bisect

#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================
#fast I/O region
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for pp in range(t):
        solve(pp)
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))

def solve(case):
    n,m = sep()
    a = lis()
    if(m<n or n<3):
        print(-1)
        return
    for i in range(n):
        a[i] = [a[i],i+1]
    a = sorted(a)
    sum = 0
    for i in range(0,n):
        sum+=2*a[i][0]
    print(sum)
    for i in range(1,n):
        print(a[i-1][1],a[i][1])

    print(a[0][1],a[n-1][1])





# testcase(1)
testcase(int(inp()))













