D. Longest Subsequence 数论

D. Longest Subsequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given array a with n elements and the number m. Consider some subsequence of a and the value of least common multiple (LCM) of its elements. Denote LCM as l. Find any longest subsequence of a with the value l ≤ m.

A subsequence of a is an array we can get by erasing some elements of a. It is allowed to erase zero or all elements.

The LCM of an empty array equals 1.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 106) — the size of the array a and the parameter from the problem statement.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of a.

Output

In the first line print two integers l and kmax (1 ≤ l ≤ m, 0 ≤ kmax ≤ n) — the value of LCM and the number of elements in optimal subsequence.

In the second line print kmax integers — the positions of the elements from the optimal subsequence in the ascending order.

Note that you can find and print any subsequence with the maximum length.

Examples

Input

Copy

7 8
6 2 9 2 7 2 3

Output

Copy

6 5
1 2 4 6 7

Input

Copy

6 4
2 2 2 3 3 3

Output

Copy

2 3
1 2 3

#include<bits/stdc++.h>
#define ll long long  
using namespace std;
ll a[1000006];
ll mp[1000006],mp2[1000006];
int main()
{
	ll n,m;
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		scanf("%lld",&a[i]);
		if(a[i]<=m)mp[a[i]]+=1;//a[i]自己对[1,m]的贡献 
	
	}
	ll ans=1;
	
	for(int i=1;i<=m;i++)
	{				
		for(int j=i;j<=m;j+=i)
		{
			mp2[j]+=mp[i];	//每个数自身倍数 对[1,m]的贡献 		
		}
	}
	for(int i=2;i<=m;i++)
	{
		if(mp2[i]>mp2[ans]) ans=i;//找出贡献最大且最小的 倍数值   即答案 
	}
	printf("%lld %lld\n",ans,mp2[ans]);
	int flag=0;
	for(int i=0;i<n;i++)
	{
		if(ans%a[i]==0)
		{			
		if(flag==0)
		{
		flag=1;	
		printf("%d",i+1);
		}
		else printf(" %d",i+1);
		}
	}
}