human
This is the easy version of the problem. The difference between the versions is that the easy version does not require you to output the numbers of the rods to be removed. You can make hacks only if all versions of the problem are solved.

Stitch likes experimenting with different machines with his friend Sparky. Today they built another machine.

The main element of this machine are n rods arranged along one straight line and numbered from 1 to n inclusive. Each of these rods must carry an electric charge quantitatively equal to either 1 or -1 (otherwise the machine will not work). Another condition for this machine to work is that the sign-variable sum of the charge on all rods must be zero.

More formally, the rods can be represented as an array of n numbers characterizing the charge: either 1 or -1. Then the condition must hold: a_1 - a_2 + a_3 - a_4 + … = 0, or ∑_{i=1}^n (-1)^{i-1} ⋅ a_i = 0.

Sparky charged all n rods with an electric current, but unfortunately it happened that the rods were not charged correctly (the sign-variable sum of the charge is not zero). The friends decided to leave only some of the rods in the machine. Sparky has q questions. In the ith question Sparky asks: if the machine consisted only of rods with numbers l_i to r_i inclusive, what minimal number of rods could be removed from the machine so that the sign-variable sum of charges on the remaining ones would be zero? Perhaps the friends got something wrong, and the sign-variable sum is already zero. In that case, you don't have to remove the rods at all.

If the number of rods is zero, we will assume that the sign-variable sum of charges is zero, that is, we can always remove all rods.

Help your friends and answer all of Sparky's questions!

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 10^3), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains two positive integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the...
# Suj Here minimal number of rods could be removed from the machine so that the sign-variable sum of charges on the remaining ones would be zero?
assistant
#Here minimal number of rods could be removed from the machine so that the sign-variable sum of charges on the remaining ones would be zero.
ith
from sys import stdin,stdout,setrecursionlimit
from io import BytesIO, IOBase
# from math import gcd,floor,sqrt,ceil
# from collections import Counter,deque
# from itertools import accumulate as acc,permutations as perm
# from bisect import bisect_left as bl,bisect_right as br,bisect as bis

setrecursionlimit(10000)

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        import os
        self.os = os
        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
        self.BUFSIZE = 8192

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

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

    def flush(self):
        if self.writable:
            self.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")

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

#...