CodeChef Little Elephant and Balance

Given an array A₁,A₂...A_N, you have to print the size of the largest contiguous subarray such that
LCM of all integers in that subarray is equal to the product of all integers in that subarray.

Formally,
For a subarray A_i,A_i+1...A_j where 1 ≤ i < j ≤ N to be valid: LCM(A_i,A_i+1...A_j) should be equal to A_i*A_i+1*...*A_j. You have to print the size of the largest valid subarray.

If no subarray satisfies output -1.

Note:A single element is not considered a subarray according to definition in this problem.

Input

First line contains T, the number of testcases. Each testcase consists of N in one line followed by N integers in next line.

Output

For each testcase, print the required answer in one line.

Constraints

• 1 ≤ T ≤ 50
• 2 ≤ N ≤ 10⁵
• 1 ≤ A_i ≤ 10⁶

Example

Input:
3
2
7 2
4
2 2 3 4
3
2 2 4

Output:
2
2
-1

Explanation

Example case 1.LCM(2,7)=2*7. Therefore, subarray A₁ to A₂ satisfies.

Example case 2.The subarrays A₂ to A₃ and A₃ to A₄ are the maximum size possible.

Example case 3.No subarray will satisfy.

Warning: Use fast input/output. Large input files. Solutions may not pass in slower languages.

Update: Time limit for python=10s

给定序列， 求最长连续序列使得 lcm( Ax, ..... Ay ) = Ax*Ax+1*....*Ay .

满足要求的时候 ， Ax ~ Ay 这些数要符合， 他们的质因子没有重复。

NlogN预处理质因子，dp出那个最右边的位置即可更新出答案 。~

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
const int M = 1000010;
int n,e[N],pos[N],ans[N],to[M],f[M];
bool not_pri[M] ;

void init() {
    int tot = 0 ;
    for( int i = 2 ; i < M ; ++i ) if( !not_pri[i] ) {
        to[i] = ++tot; f[i] = i;
        for( int j = i + i ; j < M ; j += i ){
            not_pri[j] = true ; f[j] = i;
        }
    }
}

int Work( int num , int idx ) {
    int res = 0 ;
    while( num > 1 ){
        int tmp = f[num];
        if( pos[ to[tmp] ] ) res = max( res , pos[to[tmp]] );
        pos[ to[tmp] ] = idx ;
        while( num % tmp == 0 ) num /= tmp;
    }
    return res ;
}

void Run() {
    scanf("%d",&n);
    for( int i = 1 ; i <= n ; ++i ) scanf("%d",&e[i]);
    memset( pos , 0 , sizeof pos );
    for( int i = 1 ; i <= n ; ++i ) {
        ans[i] = max( ans[i-1] , Work( e[i] , i ) );
    }
    int res = 0 ;
    for( int i = 1 ; i <= n ; ++i ) res = max( res , i - ans[i] );
    if( res <= 1 ) puts("-1");
    else printf("%d
",res);
}

int main()
{
//    freopen("in.txt","r",stdin);
    init();
    int _ , cas = 1 ;
    scanf("%d",&_);
    while(_--)Run();
}