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

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).
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static int mod=(int)1e9+7;
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        StringBuilder sb=new StringBuilder();
        int n = sc.nextInt();
        int[] arr = new int[n + 1];
        for (int i = 1; i <= n; i++)arr[i] = sc.nextInt();
        int l, r, p = 0, max = -1;
        int[] ans = new int[n];
        for(int i = 1; i <= n;) {
            l = r = i;
            while(l > 1 && arr[l - 1] % arr[i] == 0) l--;
            while(r < n && arr[r + 1] % arr[i] == 0) r++;

            i = r + 1;
            r -= l;
            if(r > max) {
                p = 0;
                max = r;
            }
            if(r == max) {
                ans[p++] = l;
            }
        }
        sb.append(p+" "+max+"\n");
        for(int i=0;i<p;i++)sb.append(ans[i]+" ");
        System.out.println(sb);
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        br = new BufferedReader(new
                InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
