A. Morning Sandwich

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Monocarp always starts his morning with a good ol' sandwich. Sandwiches Monocarp makes always consist of bread, cheese and/or ham.

A sandwich always follows the formula:

• a piece of bread
• a slice of cheese or ham
• a piece of bread
• ……
• a slice of cheese or ham
• a piece of bread

So it always has bread on top and at the bottom, and it alternates between bread and filling, where filling is a slice of either cheese or ham. Each piece of bread and each slice of cheese or ham is called a layer.

Today Monocarp woke up and discovered that he has b� pieces of bread, c� slices of cheese and hℎ slices of ham. What is the maximum number of layers his morning sandwich can have?

Input

The first line contains a single integer t� (1≤t≤10001≤�≤1000) — the number of testcases.

Each testcase consists of three integers b,c�,� and hℎ (2≤b≤1002≤�≤100; 1≤c,h≤1001≤�,ℎ≤100) — the number of pieces of bread, slices of cheese and slices of ham, respectively.

Output

For each testcase, print a single integer — the maximum number of layers Monocarp's morning sandwich can have.

Example

input

Copy

3

2 1 1

10 1 2

3 7 8

output

Copy

3
7
5

Note

In the first testcase, Monocarp can arrange a sandwich with three layers: either a piece of bread, a slice of cheese and another piece of bread, or a piece of bread, a slice of ham and another piece of bread.

In the second testcase, Monocarp has a lot of bread, but not too much filling. He can arrange a sandwich with four pieces of bread, one slice of cheese and two slices of ham.

In the third testcase, it's the opposite — Monocarp has a lot of filling, but not too much bread. He can arrange a sandwich with three pieces of bread and two slices of cheese, for example.

解题说明：此题是一道模拟题，按照题目意思做三明治，比较面包的个数和两种夹心的个数即可。

#include
int t;
int b, c, h;
int main()
{
	scanf("%d", &t);
	for (int i = 0; i < t; i++) 
	{
		scanf("%d%d%d", &b, &c, &h);
		if (b <= c + h)
		{
			printf("%d\n", b + (b - 1));
		}
		else
		{
			printf("%d\n", (c + h) + (c + h + 1));
		}
	}
	return 0;
}