D. Pair of Numbers Codeforces Round #209 (Div. 2)

D. Pair of Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Simon has an array a1, a2, ..., an, 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 al, al + 1, ..., ar are divisible by aj;
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, ..., an (1 ≤ ai ≤ 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.

Sample test(s)

input

54 6 9 3 6

output

1 32 

input

51 3 5 7 9

output

1 41 

input

52 3 5 7 11

output

5 01 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).

求出一个最长的序列，使得序列中存在一个所有数的公约数

#include <stdio.h>#include <iostream>#include <string>#include <cstring>#include <algorithm>#define N 300009using namespace std;int a[N];int n;int ans[N];int main(){    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        scanf("%d",&a[i]);        int le,ri,tmp=-1,num=0;        for(int i=1;i<=n;)        {            le=ri=i;            while(le && a[le]%a[i]==0) le--;            while(ri<=n && a[ri]%a[i]==0) ri++;            i=ri;            ri=ri-le-2;            if(ri>tmp)            {                num=0;                tmp=ri;            }            if(ri==tmp)            ans[num++]=le+1;        }        printf("%d %d\n",num,tmp);        for(int i=0;i<num;i++)        {            if(i==0)            cout<<ans[i];            else            cout<<" "<<ans[i];        }        cout<<endl;    }    return 0;}