Vasya and Coins (Codeforces 1660A)
Tags: Greedy, math

Vasya decided to go to the grocery store. He found in his wallet $$$a$$$ coins of $$$1$$$ burle and $$$b$$$ coins of $$$2$$$ burles. He does not yet know the total cost of all goods, so help him find out $$$s$$$ ($$$s > 0$$$): the minimum positive integer amount of money he cannot pay without change or pay at all using only his coins.For example, if $$$a=1$$$ and $$$b=1$$$ (he has one $$$1$$$-burle coin and one $$$2$$$-burle coin), then:  he can pay $$$1$$$ burle without change, paying with one $$$1$$$-burle coin,  he can pay $$$2$$$ burle without change, paying with one $$$2$$$-burle coin,  he can pay $$$3$$$ burle without change by paying with one $$$1$$$-burle coin and one $$$2$$$-burle coin,  he cannot pay $$$4$$$ burle without change (moreover, he cannot pay this amount at all). So for $$$a=1$$$ and $$$b=1$$$ the answer is $$$s=4$$$.

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases in the test.The description of each test case consists of one line containing two integers $$$a_i$$$ and $$$b_i$$$ ($$$0 \le a_i, b_i \le 10^8$$$) — the number of $$$1$$$-burle coins and $$$2$$$-burles coins Vasya has respectively.

Output: For each test case, on a separate line print one integer $$$s$$$ ($$$s > 0$$$): the minimum positive integer amount of money that Vasya cannot pay without change or pay at all.

Note:   The first test case of the example is clarified into the main part of the statement.  In the second test case, Vasya has only $$$1$$$ burle coins, and he can collect either any amount from $$$1$$$ to $$$4$$$, but $$$5$$$ can't.  In the second test case, Vasya has only $$$2$$$ burle coins, and he cannot pay $$$1$$$ burle without change.  In the fourth test case you don't have any coins, and he can't even pay $$$1$$$ burle. 

Sample Inputs:
5
1 1
4 0
0 2
0 0
2314 2374

Sample Outputs:
4
5
1
1
7063

Solution (Python):
for s in [*open(0)][1:]:
    (a, b) = map(int, s.split())
    print((2 * b + a) * (a > 0) + 1)

Solution (Python):
for s in [*open(0)][1:]:
    (a, b) = map(int, s.split())
    print((2 * b + a) * (a > 0) + 1)

Solution (Python):
for s in [*open(0)][1:]:
    (a, b) = map(int, s.split())
    print((2 * b + a) * (a > 0) + 1)