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.*;
import java.util.*;

public class B {

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt(), a[] = new int[n];
		for(int i = 0; i < n; ++i)
			a[i] = sc.nextInt();

		int[] L = new int[n], R = new int[n];
		// compute left bound
		Stack<Integer> s = new Stack<>();
		for(int i = 0; i < n; ++i)
		{
			while(!s.isEmpty() && a[s.peek()] % a[i] == 0)
				s.pop();
			L[i] = s.isEmpty() ? 0 : s.peek() + 1;
			s.push(i);
		}
		// compute right bound
		s = new Stack<>();
		for(int i = n - 1; i >= 0; --i)
		{
			while(!s.isEmpty() && a[s.peek()] % a[i] == 0)
				s.pop();
			R[i] = s.isEmpty() ? n - 1 : s.peek() - 1;
			s.push(i);
		}

		int maxLen = 0;
		for(int i = 0; i < n; ++i)
			maxLen = Math.max(maxLen, R[i] - L[i]);

		TreeSet<Integer> sol = new TreeSet<>();
		for(int i = 0; i < n; ++i)
			if(R[i] - L[i] == maxLen)
				sol.add(L[i] + 1);

		out.println(sol.size() + " " + maxLen);
		for(int x: sol)
			out.print(x + " ");
		out.close();
	}
	static class Scanner
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public Scanner(FileReader s) throws FileNotFoundException {	br = new BufferedReader(s);}

		public String next() throws IOException
		{
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {return br.readLine();}

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

		public boolean ready() throws IOException {return br.ready();}
	}
}