https://vjudge.net/problem/1221709/origin

解题过程有点曲折，我不赘述了，可以看CF的题解

讲几个重要的点:

1.可以用埃式筛法预处理出p[i]表示i个最小或者最大质因数(自由选择),然后分解质因数的难度就会降到logn

2.利用模板中的那片代码，可以预求出fac[1~n]的逆元，通过这种方法，就可以O(1)求出mod p   ,(i,j<=n)爽极了

代码

//Problem:
//Date:
//Skill:
//Bug:
/////////////////////////////////////////Definations/////////////////////////////////////////////////
//循环控制
#define CLR(a) memset((a),0,sizeof(a))
#define F(i,a,b) for(int i=a;i<=int(b);++i)
#define F2(i,a,b) for(int i=a;i>=int(b);--i)
#define RE(i,n)  for(int i=0;i<int(n);i++)
#define RE2(i,n) for(int i=1;i<=int(n);i++)
//输入输出
//#define INC(c) do{scanf("%c",&c);}while(isspace(c))
//#define ON cout<<endl
#define PII pair<int,int>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long llinf = 0x3f3f3f3f3f3f3f3f;
////////////////////////////////////////Options//////////////////////////////////////////////////////
typedef long long ll;
//#define stdcpph
#define CPP_IO

#ifdef stdcpph
#include<bits/stdc++.h>
#else
#include<ctype.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<algorithm>
#include<functional>
#ifdef CPP_IO
#include<iostream>
#include<iomanip>
#include<string>
#else
#include<stdio.h>
#endif
#endif
////////////////////////////////////////Basic Functions//////////////////////////////////////////////
template<typename INint>
inline void IN(INint &x)
{
	x = 0; int f = 1; char c; cin.get(c);
	while (c<'0' || c>'9') { if (c == '-')f = -1; cin.get(c); }
	while (c >= '0'&&c <= '9') { x = x * 10 + c - '0'; cin.get(c); }
	x *= f;
}
template<typename INint>
inline void OUT(INint x)
{
	if (x > 9)OUT(x / 10); cout.put(x % 10 + '0');
}
////////////////////////////////////////Added Functions//////////////////////////////////////////////
const int maxn = int(1e6+4);
const int maxnum = 25;
ll p[maxn];
ll po[maxn];
ll C[maxn][maxnum];
ll fac[maxn+maxnum], inv_fac[maxn+maxnum];
ll mod = 1e9 + 7;
void add(ll &a, const ll &b )
{
	a += b; if (a > mod)a -= mod;
}
void mul(ll &a, const ll b)
{
	a *= b; if (a > mod)a %= mod;
}
ll qpow(ll a, ll b, ll p)   //求a^bMODp
{
	ll ret = 1;
	while (b)
	{
		if (b & 1)
			ret = (ret * a) % p;
		a = (a * a) % p;
		b >>= 1;
	}
	return ret;
}
////////////////求逆元//////////////////////
////////费马小定理
ll inv(ll a, ll p)          //返回a对p的逆元
{
	return qpow(a, p - 2, p);
}
void init(ll n)
{
	p[1] = 0;
	for (ll i = 2; i*i <= n; ++i)
	{
		if (!p[i])
		{
			for (ll j = i; j <= n; j += i)
			{
				p[j] = i;
			}
		}
	}
	po[0] = 1;
	RE2(i, n)
	{
		po[i] = (po[i - 1] << 1) % mod;
	}

	fac[0] = 1;
	for (ll i = 1; i < maxn + maxnum; i++) {
		fac[i] = fac[i - 1] * i %mod;
	}
	inv_fac[maxn + maxnum - 1] = qpow(fac[maxn + maxnum - 1], mod - 2,mod);
	for (ll i = maxn + maxnum - 2; i >= 0; i--) {
		inv_fac[i] = inv_fac[i + 1] * (i + 1) % mod;
	}

}
ll solve(ll x, ll y)
{
	vector<ll>v;
	while (x!=1)
	{
		ll mm = p[x];
		if (mm)
		{
			int tm(0);
			while (x%mm == 0)
			{
				++tm; x /= mm;
			}
			v.push_back(tm);
		}
		else
		{
			v.push_back(1);
			x = 1;
		}
	}

	ll ans(1);
	for (auto p : v)
	{
		//ll inc = po[y - 1];
		ll inc = 1;
		//mul(inc, po[y - 1]);
		mul(inc, fac[y + p - 1]);
		mul(inc, inv_fac[p]);
		mul(inc, inv_fac[y - 1]);
		/*add(ans, inc);*/
		mul(ans, inc);
	}
	mul(ans, po[y - 1]);
	//mul(ans, po[y - 1]);
	//mul(ans, inv_fac[y - 1]);
	cout << ans << endl;
	return 1;
}
////////////////////////////////////////////Code/////////////////////////////////////////////////////
int main()
{
	//freopen("C:\\Users\\VULCAN\\Desktop\\data.in", "r", stdin);
	int T(1), times(0);
#ifdef CPP_IO// CPP_IO
	std::ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	cin >> T;
#else
	//IN(T);
#endif 
	/////////////////////////////////////////////////////////////////////////////////////////////////
	init(1000000 + 2);
	while (++times, T--)
	{
		ll x, y; cin >> x >> y;
		solve(x, y);
	}
	///////////////////////////////////////////////////////////////////////////////////////////////////
	return 0;
}