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题目链接：http://codeforces.com/problemset/problem/359/D点击打开链接

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.

Examples

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).

#include <iostream>#include <stdio.h>#include <set>#include <vector>#include <map>#include <algorithm>#include <string>#include <cstring>#include <limits.h>#include <math.h>#include <stack>#include <queue>using namespace std;int gcd(int x,int y){    if(x%y==0||y%x==0)        return 1;    return 0;}int a[300010];int la[300010];int ra[300010];set<int > q;set<int > ::iterator it;int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    stack<int > s;    for(int i=1;i<=n;i++)    {        la[i]=i;        if(s.empty())            {                s.push(i);            }        else        {            while(!s.empty())            {                if(a[s.top()]%a[i]==0)                {                    la[i]=la[s.top()];                    s.pop();                }                else if(a[i]%a[s.top()]==0)                {                    s.push(i);                    break;                }                else                {                    s.push(i);                    break;                }            }            if(s.empty())                s.push(i);        }    }    while(!s.empty())        s.pop();    for(int i=n;i>=1;i--)    {        ra[i]=i;        if(s.empty())            {                s.push(i);            }        else        {            while(!s.empty())            {                if(a[s.top()]%a[i]==0)                {                    ra[i]=ra[s.top()];                    s.pop();                }                else if(a[i]%a[s.top()]==0)                {                    s.push(i);                    break;                }                else                {                    s.push(i);                    break;                }            }            if(s.empty())                s.push(i);        }    }    int maxn=0;int cnt=0;    for(int i=1;i<=n;i++)        maxn=max(maxn,ra[i]-la[i]);    for(int i=1;i<=n;)        if((ra[i]-la[i])==maxn)            cnt++,q.insert(la[i]),i=(ra[i]+1);        else            i++;    printf("%d %d\n",cnt,maxn);    for(it=q.begin();it!=q.end();it++)        printf("%d ",(*it));}