cf359D Pair of Numbers

Simon has an array a₁, a₂, ..., a_n, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:

1. there is integer j (l ≤ j ≤ r), such that all integers a_l, a_l + 1, ..., a_r are divisible by a_j;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;

Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵).

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁶).

Output

Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.

Example

Input

5
4 6 9 3 6

Output

1 3
2 

Input

5
1 3 5 7 9

Output

1 4
1 

Input

5
2 3 5 7 11

Output

5 0
1 2 3 4 5 

Note

In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.

In the second sample all numbers are divisible by number 1.

In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).

 

要找个最长区间，使得区间里存在一个数x，区间其他数都能被它整除

对于一个i，直接暴力O(n)找左右两边能被a[i]整除的最远的l,r

同时，如果找到了一个i的[l,r]，那么对于所有i<j<=r，a[i] | a[j]，所以a[j]的对应[l,r]不会比i的更优

所以做完 i 之后可以直接 i 跳到 r+1 的位置

这样还挺快啊？

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
int a[300010];
int len,tot;
int ans[300010];
int main()
{
    n=read();
    for (int i=1;i<=n;i++)a[i]=read();
    for (int i=1;i<=n;i++)
    {
        int l=i,r=i;
        while (l>=1&&a[l]%a[i]==0)l--;l++;
        while (r<=n&&a[r]%a[i]==0)r++;r--;
        if (r-l>len){len=r-l;tot=1;ans[tot]=l;}
        else if (r-l==len)ans[++tot]=l;
        i=r;
    }
    printf("%d %d\n",tot,len);
    for(int i=1;i<=tot;i++)printf("%d ",ans[i]);
}