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.Scanner;

public class Main {
	static int[] a;
	static int n;
	public static int processDP (int k, int num, int delta, int[] dp) {
		if (k >= n || k < 0) return 0;
		if (a[k] % num == 0) {
			return 1 + dp[k] + processDP(k + delta * (dp[k] + 1), num, delta, dp);
		} else {
			return 0;
		}
	}
	public static void main (String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		a  = new int[n];
		int[] dpr = new int[n];
		int[] dpl = new int[n];
		int[] dp = new int[n];
		dpr[n-1] = 0;
		dpl[0] = 0;
		for (int i = 0; i < a.length; i++) {
			a[i] = in.nextInt();
		}
		for (int i = 1; i < n; i++) {
			dpl[i] = processDP(i-1, a[i], -1, dpl);
		}
		for (int i = n-2; i >= 0; i--) {
			dpr[i] = processDP(i+1, a[i], 1, dpr);
		}
		int maxi = -1;
		for (int i = 0; i < n; i++) {
			dp[i] = dpl[i] + dpr[i];
			if (dp[i] > maxi) {
				maxi = dp[i];
			}
		}
		StringBuilder s = new StringBuilder();
		boolean first = true;
		int prev = -1;
		int num = 0;
		for (int i = 0; i < n; i++) {
			int index = i-dpl[i]+1;
			if (dp[i] == maxi && index != prev) {
				if (!first) {
					s.append(" ");
				}
				num++;
				s.append(index);
				prev = index;
				first = false;
			}
		}
		System.out.println(num + " " + maxi);
		System.out.println(s);
		in.close();
	}
}