AashishNandakumar

#User function Template for python3
from collections import deque

class Solution:
	def addBinary(self, A, B):
	# code here
        len1 = len(A)
        len2 = len(B)
            
        carry = 0

AashishNandakumar

# Holloumi Boxes


# logic
def can_sort_it(a, n, k) -> bool:
    # def reverse(a, ptr1, ptr2):
    #     a[ptr1:ptr2] = reversed(a[ptr1:ptr2])

    # if n == 1:
    #     return True

AashishNandakumar

# logic
def can_take_the_trip() -> int:
    """
    logic:
    calculate gaps between each petrol stations, the max gap is the minimum fuel(1unit = 1lt) needed to complete the trip
    """
    n, x = map(int, input().split())
    petrol_stations = list(map(int, input().split()))

    # approach 1

AashishNandakumar

# Cover in Water
import re


def cover_in_water():
    n = int(input())
    s = input()

    # logic here

AashishNandakumar

def game_with_integer() -> str:
    n = int(input())
    if n % 3 == 0:
        return "Second"
    else:
        return "First"


t = int(input())
for _ in range(t):

AashishNandakumar

class Solution:
    def isHappy(self, n: int) -> bool:
        def get_sum_of_squares_of_digits(n) -> int:
            res = 0
            while n != 0:
                res += (n % 10) ** 2
                n //= 10
            return res

        visited = set()

AashishNandakumar

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        n = len(nums)

        # approach-1: brute force
        # for i in range(n-1):
        #     for j in range(i+1, n):
        #         if abs(i-j) <= k and nums[i] == nums[j]:
        #             return True

AashishNandakumar

def jagged_swaps():
    n = int(input())
    a = list(map(int, input().split()))

    if a[0] == 1:
        return "YES"
    return "NO"

    """Actual sorting logic if a[0] == 1"""
    # upper_bound = n

AashishNandakumar

def doremys_paint():
    n = int(input())
    a = list(map(int, input().split()))

    # logic here
    """
    case1: if there is only 1 unique element, return 'YES'
    case2: if there is 2 unique elements, and if any one element's count/frequency == n/2,
        return 'YES' or 'NO' otherwise
    case 3: if there is more than 2 unique elements return 'NO'

AashishNandakumar

def dont_try_to_count():
    n, m = map(int, input().split())
    x = input()
    s = input()

    """Approach 1"""
    # operations = 0
    # while len(x) <= n * m:
    #     if s in x:
    #         return operations