【CODECHEF】【phollard rho + miller_rabin】The First Cube

All submissions for this problem are available.

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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 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 ;
 }