Codeforces 893E - Counting Arrays 【组合数】

E. Counting Arrays

time limit per test  3  seconds

memory limit per test      256 megabytes

You are given two positive integer numbers x and y. An array F is called an y-factorization of x iff the following conditions are met:

·   There are y elements in F, and all of them are integer numbers;

·   .

You have to count the number of pairwise distinct arrays thatare y-factorizations of x. Two arrays A and B are considered different iff there existsat least one index i (1 ≤ i ≤ y) such that A_i ≠ B_i. Since the answer can bevery large, print it modulo 10⁹ + 7.

Input

The first line contains one integer q (1 ≤ q ≤ 10⁵) — the number of testcasesto solve.

Then q lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ 10⁶). Each of these linesrepresents a testcase.

Output

Print q integers. i-th integer has to be equal to the number of y_i-factorizations of x_i modulo 10⁹ + 7.

Example

Input

2
6 3
4 2

Output

36
6

Note

In the second testcase of the example there are six y-factorizations:

·   { - 4,  - 1};

·   { - 2,  - 2};

·   { - 1,  - 4};

·   {1, 4};

·   {2, 2};

·   {4, 1}.

 

【题意】

给出两个数x、y，问有多少种排列F₁,F₂,F₃……Fn，使得它们的乘积为y，答案模1e9+7。

【思路】

我们首先对x分解质因数，分解成p1^c1+ p2^c2……的形式。

然后先不考虑符号，把排列中的数Fi都置为1，然后依次处理每个底数，假设它的指数为C，那么这里贡献的次数便是将C个底数分配到y个位置的情况数目。

转化一下，便是放球问题中的球相同，盒子不同，允许空盒的方案数，具体可以看这份博客：传送门

然后每个底数的贡献相乘即可。

最后再考虑符号，我们很容易发现，其实对于y个数的情况，前y-1个数的符号可以任意取定，只要利用最后一个数的符号使得乘积为正即可。

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 2000005;
const ll mod = 1e9+7;
const ll INF = 1e15;
const double eps = 1e-9;

ll fac[maxn];
ll inv[maxn];

ll fast_mod(ll a,ll n,ll Mod)
{
    ll ans=1;
    a%=Mod;
    while(n)
    {
        if(n&1) ans=(ans*a)%Mod;
        a=(a*a)%Mod;
        n>>=1;
    }
    return ans;
}

void init()
{
    fac[0]=1;
    for(int i=1;i=0;i--)
    {
        inv[i]=(inv[i+1]*(i+1))%mod;
    }
}

ll C(int n,int m)
{
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}

int main()
{
    init();
    rush()
    {
        int x,y;
        scanf("%d%d",&x,&y);
        vectorvec;
        for(int i=2;i*i<=x;i++)     //分解质因数
        {
            if(x%i==0)
            {
                int cnt=0;
                while(x%i==0)
                {
                    cnt++;
                    x/=i;
                }
                vec.push_back(cnt);
            }
        }
        if(x>1) vec.push_back(1);
        ll ans=1;
        y--;
        for(int i=0;i