Educational Codeforces Round 9-D. Longest Subsequence(筛选法)

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

7 86 2 9 2 7 2 3

output

6 51 2 4 6 7

input

6 42 2 2 3 3 3

output

2 31 2 3

题意：
  给你n个数，要你找到最小公倍数在《=m的最大个数
  
思路：
   因为数据较大，所以暴力肯定超时的。这题可以用筛选法来做，
   因为只有倍数关系才能符合题意。 

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define inf 0x3f3f3f3f#define T 1000010typedef long long ll;int d[T],a[T],c[T];int main(){int n,m,i,j,k;while(~scanf("%d%d",&n,&m)){fill(c,c+T,0);fill(d,d+T,0);for(i=0;i<n;++i){scanf("%d",&a[i]);if(a[i]<=m){//记录小于m的数出现的次数 c[a[i]]++;}}for(i=m;i;--i){//在【m，0】范围内的数 for(j=i;j<=m;j+=i){//将【i，m】内的数推上上面 d[j] += c[i];}}ll ans1,ans2;//记录答案1，答案2 ans1 = ans2 = -1;//赋值为最小 for(i=1;i<=m;++i){//在【m，0】内找一个最大的个数 if(d[i]>ans1){ans1 = d[i];ans2 = i;}}printf("%I64d %I64d\n",ans2,ans1);for(i=0;i<n;++i){//输出符合的数 if(ans2%a[i]==0)printf("%d ",i+1);}printf("\n");}return 0;}