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.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
	private static Scanner cin;
	private static int N = 0;
	private static int res = -1;
	private static ArrayList<Integer> solutions = new ArrayList<Integer>();
	private static int lst[] = new int[300010];
	private static int M[][] = new int[300010][20];
	private static int G[][] = new int[300010][20]; 

	private static int gcd(int x, int y){
		if (y == 0){
			return x;
		}
		if (x % y == 0){
			return y;
		}else{
			return gcd(y, x % y);
		}
	}
	
	private static long getG(int start, int end){
		int len = end - start + 1;
		
		int k = 0, tmp = len;
		while (tmp > 0){
			tmp /= 2;
			k++;
		}
		k--;
		
		if ((1<<k) == len){
			return G[start][k];
		}else{
			return gcd(G[start][k], G[end - (1<<k) + 1][k]);
		}
	}
	
	private static long getM(int start, int end){
		int len = end - start + 1;
		int k = 0, tmp = len;
		while (tmp > 0){
			tmp /= 2;
			k++;
		}
		k--;
		
		if ((1<<k) == len){
			return M[start][k];
		}else{
			return Math.min(M[start][k], M[end - (1<<k) + 1][k]);
		}
	}
	
	private static boolean check(int length){
		boolean ret = false;
		
		for (int i = 0; i + length - 1 < N; i++){
			long g = getG(i, i + length - 1);
			long m = getM(i, i + length - 1);
			
			if (g == m){
				if (length > res){
					res = length;
					solutions.clear();
					solutions.add(i + 1);
				}else if (length == res){
					solutions.add(i + 1);
				}
				ret = true;
			}
		}
		return ret;
	}
	public static void main(String[] args) {		
		cin = new Scanner(System.in);
		
		N = cin.nextInt();
		for (int i = 0; i < N; i++){
			lst[i] = cin.nextInt();
		}
		
		int bound = 0, sz = 1;
		while (sz < N){
			sz += sz;
			bound++;
		}
		
		for (int i = 0; i < N; i++){
			M[i][0] = lst[i];
			G[i][0] = lst[i];
		}
		
		for (int j = 1; j <= bound; j++){
			for (int i = 0; i + (1<<j) - 1 < N; i++){
				G[i][j] = gcd(G[i][j - 1], G[i + (1<<(j - 1))][j - 1]);
				
				if (M[i][j - 1] <= M[i + (1<<(j - 1))][j - 1]){
					M[i][j] = M[i][j - 1];
				}else{
					M[i][j] = M[i + (1<<(j - 1))][j - 1];
				}
			}
		}
		
		int up = N, down = 1;
		while (up > down){
			int mid = (up + down + 1) / 2;
			
			if (check(mid)){
				down = mid;
			}else{
				up = mid - 1;
			}
		}
		
		if (res == -1){
			check((up + down) / 2);
		}
		
		res--;
		System.out.println(solutions.size() + " " + res);
		
		Collections.sort(solutions);
		System.out.print(solutions.get(0));
		for (int i = 1; i < solutions.size(); i++){
			System.out.print(" " + solutions.get(i));
		}
		System.out.print("\n");
	}

}
