【CODECHEF】【phollard rho + miller

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Read problems statements in Mandarin Chinese and Russian.

This problem's statement is really a short one.

You are given an integer S. Consider an infinite sequence S, 2S, 3S, ... . Find the first number in this sequence that can be represented as Q³, where Q is some positive integer number. As the sought number could be very large, please print modulo (10⁹ + 7).

The number S will be given to you as a product of N positive integer numbers A₁, A₂, ..., A_N, namely S = A₁ * A₂ * ... * A_N

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.

The first line of each test case contains an integer N.

Then there is a line, containing N space separated integers, denoting the numbers A₁, A₂, ..., A_N.

Output

For each test case, output a single line containing the first term of the sequence which is the perfect cube, modulo 10⁹+7.

Constraints

1 ≤ T ≤ 10
(Subtask 1): N = 1, 1 ≤ S ≤ 10⁹ - 15 points.
(Subtask 2): N = 1, 1 ≤ S ≤ 10¹⁸ - 31 point.
(Subtask 3): 1 ≤ N ≤ 100, 1 ≤ A_i ≤ 10¹⁸ - 54 points.

Example

Input:

2

2

2 2

2

2 3

Output:

8

216

Explanation

Example case 1. First few numbers in the infinite sequence 4, 8, 12, 16, 20, , etc. In this sequence, 8 is the first number which is also a cube (as 2³ = 8).

Example case 2. First few numbers in the infinite sequence 6, 12, 18, 24, , etc. In this sequence, 216 is the first number which is also a cube (as 6³ = 216).

【分析】

挺模板的东西，就当复习一下了。

 /*

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 才过清明，渐觉伤春暮。

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 桃杏依稀香暗渡。

 谁在秋千，笑里轻轻语。

 一寸相思千万绪。人间没个安排处。

 */

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <cmath>

 #include <queue>

 #include <vector>

 #include <iostream>

 #include <string>

 #include <ctime>

 #include <map>

 #define LOCAL

 const int MAXN =  + ;

 const long long MOD = ;

 const double Pi = acos(-1.0);

 long long G = ;//原根

 const int MAXM =  *  + ;

 using namespace std;

 typedef long long ll;

 ll read(){

    ll flag = , x = ;

    char ch;

    ch = getchar();

    while (ch < '' || ch > '') {if (ch == '-') flag = -; ch = getchar();}

    while (ch >= '' && ch <= '') {x = x *  + (ch - ''); ch = getchar();}

    return x * flag;

 }

 map<ll, int>Num;//记录个数

 ll data[MAXN];

 ll n;

 ll mul(ll a, ll b, ll c){//又要用快速乘QAQ

    if (b == ) return 0ll;

    if (b == ) return a % c;

    ll tmp = mul(a, b / , c);

    if (b %  == ) return (tmp + tmp) % c;

    else return (((tmp + tmp) % c) + (a % c)) % c;

 }

 ll pow(ll a, ll b, ll c){

    if (b == ) return 1ll;

    if (b == ) return a % c;

    ll tmp = pow(a, b / , c);

    if (b %  == ) return mul(tmp, tmp, c);

    else return mul(mul(tmp, tmp, c), a, c);

 }

 bool Sec_check(ll a, ll b, ll c){

      ll tmp = pow(a, b, c);

      if (tmp !=  && tmp != (c - )) return ;

      if (tmp == (c - ) || (b %  != )) return ;

      return Sec_check(a, b / , c);

 }

 //判断n是否是素数

 bool miller_rabin(ll n){

      int cnt = ;

      while (cnt--){

            ll a = (ll)rand() % (n - ) + ;

            if (!Sec_check(a, n - , n)) return ;

      }

      return ;

 }

 ll gcd(ll a, ll b){return b == 0ll ? a : gcd(b, a % b);}

 ll pollard_rho(ll a, ll c){

    ll i = , k = ;

    ll x, y, d;

    x = (ll)((double)(rand() / RAND_MAX) * (a - ) + 0.5) + 1ll;

    y = x;

    while (){

          i++;

          x = (mul(x, x, a) % a + c) % a;

          d = gcd(y - x + a, a);

          if (d >  && d < a) return d;

          if (y == x) return a;//失败

          if (i == k){

             k <<= ;

             y = x;

          }

    }

 }

 void find(ll a, ll c){

      if (a == ) return;

      if (miller_rabin(a)){

         Num[a]++;

         return;

      }

      ll p = a;

      while (p >= a) pollard_rho(a, c--);

      pollard_rho(p, c);

      pollard_rho(a / p, c);

 }

 void init(){

      Num.clear();

      scanf("%d", &n);

      for (int i = ; i <= n; i++) {

          data[i] = read();

          find(data[i], );

      }

 }

 void work(){

      ll Ans = ;

      for (int i = ; i <= n; i++) Ans = (Ans * data[i]) % MOD;

      for (map<ll, int>::iterator it = Num.begin(); it != Num.end(); it++){

          it->second %= ;

          if (it->second){

             for (int i = it->second; i < ; i++) Ans = (Ans * ((it->first) % MOD)) % MOD;

          }

      }

      printf("%lld\n", Ans);

 }

 int main(){

     int T;

     scanf("%d", &T);

     while (T--){

           init();

           work();

     }

     return ;

 }