There is an infinite 2-dimensional grid. The robot stands in cell (0, 0) and wants to reach cell (x, y). Here is a list of possible commands the robot can execute:

  * move north from cell (i, j) to (i, j + 1); 
  * move east from cell (i, j) to (i + 1, j); 
  * move south from cell (i, j) to (i, j - 1); 
  * move west from cell (i, j) to (i - 1, j); 
  * stay in cell (i, j). 



The robot wants to reach cell (x, y) in as few commands as possible. However, he can't execute the same command two or more times in a row.

What is the minimum number of commands required to reach (x, y) from (0, 0)?

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.

Each of the next t lines contains two integers x and y (0 ≤ x, y ≤ 10^4) — the destination coordinates of the robot.

Output

For each testcase print a single integer — the minimum number of commands required for the robot to reach (x, y) from (0, 0) if no command is allowed to be executed two or more times in a row.

Example

Input


5
5 5
3 4
7 1
0 0
2 0


Output


10
7
13
0
3

Note

The explanations for the example test:

We use characters N, E, S, W and 0 to denote going north, going east, going south, going west and staying in the current cell, respectively.

In the first test case, the robot can use the following sequence: NENENENENE.

In the second test case, the robot can use the following sequence: NENENEN.

In the third test case, the robot can use the following sequence: ESENENE0ENESE.

In the fourth test case, the robot doesn't need to go anywhere at all.

In the fifth test case, the robot can use the following sequence: E0E.
from __future__ import division, print_function


def main():
    # Template 1.0
    import sys, re, math
    import sys
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    from itertools import permutations
    def STR():
        return list(input())

    def INT():
        return int(input())

    def MAP():
        return map(int, input().split())

    def LIST():
        return list(map(int, input().split()))

    def list2d(a, b, c):
        return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):
        return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))[::-1]
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    t = INT()

    while (t != 0):
        x,y = MAP()

        if(x==y):
            print(x+y)
        else:
            print(x+y+abs(x-y)-1)

        t -= 1


######## Python 2 and 3 footer by Pajenegod and c1729
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    main()