Codeforces 893E

E. Counting Arrays

time limit per test 3  seconds

memory limit per test     256 megabytes

You are given two positive integer numbersx andy. An arrayF is called any-factorization ofx iff the following conditions are met:

·   There arey elements inF, and all of them are integer numbers;

·   .

You have to count the number of pairwise distinct arrays thatarey-factorizations ofx. Two arraysA andB are considered different iff there existsat least one indexi (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 integerq (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 ofy_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 sixy-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 <cstdio>#include <cmath>#include <vector>#include <cstring>#include <algorithm>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<maxn;i++)    {        fac[i]=(fac[i-1]*i)%mod;    }    inv[maxn-1]=fast_mod(fac[maxn-1],mod-2,mod);    for(int i=maxn-2;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);        vector<int>vec;        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<vec.size();i++)        {            ans*=C(vec[i]+y,y);                 //模拟放球            ans%=mod;        }        ans=(ans*fast_mod(2,y,mod))%mod;        //处理符号        printf("%I64d\n",ans);    }}