A. Sum of Odd Integers

time limit per test ：2 seconds memory limit per test ：256 megabytes

input ：standard input output ：standard output

You are given two integers n and k. Your task is to find if n can be represented as a sum of k distinct positive odd (not divisible by 2) integers or not.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t ( $1\leqslant t \leqslant 10^{5}$ ) — the number of test cases.

The next t lines describe test cases. The only line of the test case contains two integers n and k ( $1\leqslant n, k\leqslant 10^7$ ).

Output

For each test case, print the answer — "YES" (without quotes) if nn can be represented as a sum of k distinct positive odd (not divisible by 2) integers and "NO" otherwise.

Example

input

output

YES
YES
NO
YES
YES
NO

Note

In the first test case, you can represent 3 as 3.

In the second test case, the only way to represent 4 is 1+3.

In the third test case, you cannot represent 10 as the sum of three distinct positive odd integers.

In the fourth test case, you can represent 10 as 3+7, for example.

In the fifth test case, you can represent 16as 1+3+5+7.

In the sixth test case, you cannot represent 16as the sum of five distinct positive odd integers.

题目大意

给你n和k，问你n能否由k个互不相同的奇数组成

思路

思路很简单（虽然很简单，但第一次看错题目，第二次没开long long wa了两发，难受！所以还是写一篇记录一下）

1. 前k个奇数的和是 $k^2$ ，如果n比 $k^2$ 小一定不可能

2. 偶数个奇数一定是偶数，奇数个奇数一定是奇数，满足 $n\geqslant k^2$ 时判断奇偶性即可，n，k奇偶性相同一定可以（为什么一定可以，我们只要取最小的k个互不相同的奇数，然后把最后一个一直加2即可，这样就能保证k个奇数仍然互不相同）

3. 注意开long long，k平方会爆

代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5 + 10;
int main()
{
    int t;
    cin >>t;
    while (t --){
        ll n, k;
        cin >> n >>k;
        if (n < k * k) printf("NO\n");
        else {
            if ((n % 2 == 1 && k % 2 == 0) || (n % 2 == 0 && k % 2 == 1)) printf("NO\n");
            else printf("YES\n");
        }
    }
    return 0;
}

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