A

A - Pair of Numbers

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:  System Crawler  (2013-11-04)

Description

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.

Sample Input

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 

 

const int maxn = 310000;
int num[maxn];
int ans[maxn];
int main() 
{
    //freopen("in.txt","r",stdin);
    int n;
    while(cin>>n)
    {
        repf(i,1,n) scanf("%d",&num[i]);
        int Max = 0;
        int cnt = 0;
        for(int i = 1;i<=n;)
        {
            int L = i;
            int R = i;
            while(L>=1 && num[L]%num[i] == 0) L--;
            while(R<=n && num[R]%num[i] == 0) R++;
            i = R;
            R--;L++;
            if(R - L > Max)
            {
                Max = R - L;
                cnt = 0;
                ans[cnt] = L;
                cnt++;
            }
            if(R - L == Max)
            {
                ans[cnt] = L;
                cnt++;
            }
        }
        int sz = unique(ans,ans+cnt) - ans;
        printf("%d %d
",sz,Max);
        printf("%d",ans[0]);
        rep(i,1,sz) printf(" %d",ans[i]);
        cout<<endl;
    }
    return 0;
}