（Educational Codeforces Round 9）Longest Subsequence（dp）

Longest Subsequence

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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

7 8
6 2 9 2 7 2 3

output

6 5
1 2 4 6 7

input

6 4
2 2 2 3 3 3

output

2 3
1 2 3

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题意
给n个数，然后你要找到一个最长的序列，使得序列中的数的lcm小于m
题解：
lcm和顺序无关，只要统计每个数有多少个。
然后再类似筛法一样，去筛每一个数的因子有多少个。

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#include <bits/stdc++.h>using namespace std;int n , m;int a[1000001] , num[1000001];int divnum[1000001];int main(){    scanf("%d%d" , &n , &m);    for(int i = 1 ; i <= n ; i++)    {        scanf("%d" , &a[i]);        if(a[i] <= m)        num[a[i]]++;    }    for(int i = 1 ; i <= m ; i++)        if(num[i])            for(int j = i ; j <= m ; j += i)                divnum[j] += num[i];    int maxi = 0;    for(int i = 1 ; i <= m ; i++)        if(divnum[i] > divnum[maxi])            maxi = i;    if(maxi == 0)    {    printf("1 0\n\n");    return 0;    }    printf("%d %d\n" , maxi , divnum[maxi]);    for(int i = 1 ; i <= n ; i++)        if(maxi % a[i] == 0)            printf("%d " , i);    printf("\n");    return 0;}