B. Robin Hood and the Major Oak

time limit per test

1 second

memory limit per test

256 megabytes

In Sherwood, the trees are our shelter, and we are all children of the forest.

The Major Oak in Sherwood is known for its majestic foliage, which provided shelter to Robin Hood and his band of merry men and women.

The Major Oak grows iiii new leaves in the ii-th year. It starts with 11 leaf in year 11.

Leaves last for kk years on the tree. In other words, leaves grown in year ii last between years ii and i+k−1i+k−1 inclusive.

Robin considers even numbers lucky. Help Robin determine whether the Major Oak will have an even number of leaves in year nn.

Input

The first line of the input contains a single integer tt (1≤t≤1041≤t≤104) --- the number of test cases.

Each test case consists of two integers nn, kk (1≤n≤1091≤n≤109, 1≤k≤n1≤k≤n) --- the requested year and the number of years during which the leaves remain.

Output

For each test case, output one line, "YES" if in year nn the Major Oak will have an even number of leaves and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Example

Input

Copy

复制代码

5

1 1

2 1

2 2

3 2

4 4

Output

Copy

复制代码
NO
YES
NO
NO
YES

Note

In the first test case, there is only 11 leaf.

In the second test case, k=1k=1, so in the 22-nd year there will be 22=422=4 leaves.

In the third test case, k=2k=2, so in the 22-nd year there will be 1+22=51+22=5 leaves.

In the fourth test case, k=2k=2, so in the 33-rd year there will be 22+33=4+27=3122+33=4+27=31 leaves.

解题说明:此题是一道数学题,每年能生成i^i个树叶,每个树叶能存活k年,求最后的数量是否是偶数。找规律能发现此题为找出以n结尾的连续整数k的和是否为偶数。

cpp 复制代码
#include<iostream>
#include<cmath>
using namespace std;

void solve()
{
	int n, k;
	cin >> n >> k;
	if (n % 2 == 0)
	{
		if (k / 2 % 2 == 0)
		{
			cout << "YES" << '\n';
			return;
		}
		else 
		{
			cout << "NO" << '\n';
			return;
		}
	}
	else
	{
		if ((k / 2 + (k % 2 != 0)) % 2 == 0)
		{
			cout << "YES" << '\n';
			return;
		}
		else
		{
			cout << "NO" << '\n';
			return;
		}
	}
}

int main() 
{
	int t = 1;
	cin >> t;
	while (t--)
	{
		solve();
	}
	return 0;
}
相关推荐
感哥15 小时前
C++ lambda 匿名函数
c++
沐怡旸21 小时前
【底层机制】std::unique_ptr 解决的痛点?是什么?如何实现?怎么正确使用?
c++·面试
感哥21 小时前
C++ 内存管理
c++
聚客AI21 小时前
🙋‍♀️Transformer训练与推理全流程:从输入处理到输出生成
人工智能·算法·llm
大怪v1 天前
前端:人工智能?我也会啊!来个花活,😎😎😎“自动驾驶”整起!
前端·javascript·算法
惯导马工1 天前
【论文导读】ORB-SLAM3:An Accurate Open-Source Library for Visual, Visual-Inertial and
深度学习·算法
骑自行车的码农1 天前
【React用到的一些算法】游标和栈
算法·react.js
博笙困了1 天前
AcWing学习——双指针算法
c++·算法
感哥1 天前
C++ 指针和引用
c++
moonlifesudo1 天前
322:零钱兑换(三种方法)
算法