893E

E. Counting Arrays

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 that are y-factorizations of x. Two arrays A and B are considered different iff there exists at least one index i (1 ≤ i ≤ y) such that Ai ≠ Bi. Since the answer can be very large, print it modulo 109 + 7.

Input

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

Then q lines follow, each containing two integers xi and yi (1 ≤ xi, yi ≤ 106). Each of these lines represents a testcase.

Output

Print q integers. i-th integer has to be equal to the number of yi-factorizations of xi 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x,y。求满足含有y个元素的之积是x的数列个数。

思路：排列组合，插空法。

代码：

#include<bits/stdc++.h>
#define db double
#define ll long long
#define vec vector<ll>
#define Mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define debug puts("-------------");
const int N = 1e6 + 2015;
const int mod = 1e9 + 7;
const int MOD = mod-1;
const db  eps = 1e-18;
const db  PI = acos(-1.0);
using namespace std;
bool v[N];
int pri[N];
ll F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘

int p=0;
void init()
{
    memset(v,0,sizeof(v));
    for(int i=2;i<1002002;i++){
        if(!v[i]) pri[p++]=i;
        for(int j=2*i;j<1002002;j+=i) {v[j]=1;}
    }
    inv[1] = 1;
    for(int i = 2; i < 1002005; i ++){
        inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
    }
    F[0] = Finv[0] = 1;
    for(int i = 1; i < 1002005; i ++){
        F[i] = F[i-1] * 1ll * i % mod;
        Finv[i] = Finv[i-1] * 1ll* inv[i] % mod;
    }
}
ll C(ll n, ll m){    //comb(n, m)就是C(n, m)
    if(m < 0 || m > n)  return 0;
    return F[n] * 1ll * Finv[n - m] % mod * Finv[m] % mod;
}
ll qpow(ll x,ll n)
{
    ll ans=1;
    x%=mod;
    while(n){
        if(n&1) ans=ans*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return  ans;
}

int main(){
    int q;
    ci(q);
    init();
    for(int i=0;i<q;i++)
    {
        ll x,y;
        cl(x),cl(y);
        ll ans=qpow(2,y-1);
        if(x==1){
            pl(qpow(2, y - 1));
            continue;
        }
        vector<int> e;e.clear();
        map<int,int> mp;mp.clear();
        int id=0;
        while(x>1){
            if(x%pri[id]==0){
                int n=0;
                while(x%pri[id]==0) n++,x/=pri[id];
                ans=ans*C(n+y-1,y-1)%mod;
            }
            id++;
            if(pri[id]>1000){
                if(x>1) ans=ans*y%mod;
                break;
            }
        }
        pl(ans);
    }
    return 0;
}