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
qq_529835358 分钟前
ThreadLocal内存泄漏 强引用vs弱引用
java·开发语言·jvm
景彡先生12 分钟前
C++并行计算:OpenMP与MPI全解析
开发语言·c++
归去_来兮2 小时前
深度学习模型在C++平台的部署
c++·深度学习·模型部署
量子联盟2 小时前
原创-基于 PHP 和 MySQL 的证书管理系统,免费开源
开发语言·mysql·php
pay4fun3 小时前
2048-控制台版本
c++·学习
时来天地皆同力.3 小时前
Java面试基础:概念
java·开发语言·jvm
hackchen3 小时前
Go与JS无缝协作:Goja引擎实战之错误处理最佳实践
开发语言·javascript·golang
YuTaoShao4 小时前
【LeetCode 热题 100】141. 环形链表——快慢指针
java·算法·leetcode·链表
hjjdebug4 小时前
ffplay6 播放器关键技术点分析 1/2
c++·ffmpeg·音视频
铲子Zzz5 小时前
Java使用接口AES进行加密+微信小程序接收解密
java·开发语言·微信小程序