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.InputStreamReader;
import java.io.PrintWriter;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class PairOfNumbers {
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);

		int N = Integer.parseInt(in.readLine());
		StringTokenizer tk = new StringTokenizer(in.readLine());
		int[] arr = new int[N];
		for (int i = 0; i < N; i++) {
			arr[i] = Integer.parseInt(tk.nextToken());
		}

		// dp[i] = j -> j = left most such that j->i are div by i
		int[] dpLeft = new int[N];
		dpLeft[0] = 0;
		for (int i = 1; i < N; i++) {
			dpLeft[i] = i;
			while (dpLeft[i] - 1 >= 0 && arr[dpLeft[i] - 1] % arr[i] == 0)
				dpLeft[i] = dpLeft[dpLeft[i] - 1];
		}

		int[] dpRight = new int[N];
		dpRight[N - 1] = N - 1;
		for (int i = N - 2; i >= 0; i--) {
			dpRight[i] = i;
			while (dpRight[i] + 1 < N && arr[dpRight[i] + 1] % arr[i] == 0)
				dpRight[i] = dpRight[dpRight[i] + 1];
		}
		
		int max = 0;
		for (int i = 0; i < N; i++) {
			int val = dpRight[i] - dpLeft[i];
			if (val > max)
				max = val;
		}

		Set<Integer> pairs = new TreeSet<Integer>();
		for (int i = 0; i < N; i++) {
			int val = dpRight[i] - dpLeft[i];
			if(max == val){
				pairs.add(dpLeft[i]+1);
			}
		}

		out.println(pairs.size() + " " + max);
		for(int i : pairs){
			out.print(i + " ");
		}
		out.close();
	}
}
