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) {
        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        /*
        Turns out there's this cool thing called a sparse table.
        It's pretty similar to binary lifting for LCA and you can use it for queries
        on a range as long as there are two properties:
        1) the data in the array doesn't change (otherwise segment tree would probably be better)
        2) the function is associative (in this case gcd(a, gcd(b,c)) = gcd(gcd(a,b),c) )
        It's much easier to implement than a segment tree and works in log n per query.

        For this problem, for each i you can binary search how far left and how far right you can go so that
        the gcd in that range is equal to a[i]. This way it's guaranteed that a[i] can divide everything in
        that range (notice that for a valid range this also has to always be the smallest value in its range)
        So binary search the length of the largest segment where a[i] is the gcd of everything and then get
        the answer from that.

        Either my computer's going slow or this solution might still time out.

         */


        int n = fs.nextInt();
        int[] a = fs.nextIntArray(n);
        int[][] sparse = new int[n][20];
        for(int i = 0; i < n; i++)
            sparse[i][0] = a[i];

        for(int pow = 1; pow < sparse[0].length; pow++) {
            int add = (1 << pow-1);
            for(int i = 0; i <= n - (1 << pow); i++) {
                sparse[i][pow] = gcd (sparse[i][pow-1], sparse[i + add][pow-1]);
            }
        }

        int[] range = new int[n];
        int max = 0;

        for(int i = 0; i < n; i++) {

            int goRight = i;
            int lo = i, hi = n-1;
            for(int bs = 0; bs < 20 && lo <= hi; bs++) {
                int mid = (lo + hi) >> 1;
                int gcd = getRange(i, mid, sparse);
                if(gcd == a[i]) {
                    lo = mid+1;
                    goRight = Math.max(goRight, mid);
                } else {
                    hi = mid-1;
                }
            }

            int goLeft = i;
            lo = 0; hi = i;
            for(int bs = 0; bs < 20 && lo <= hi; bs++) {
                int mid = (lo + hi) >> 1;
                int gcd = getRange(mid, i, sparse);
                if(gcd == a[i]) {
                    hi = mid-1;
                    goLeft = Math.min(goLeft, mid);
                } else {
                    lo = mid+1;
                }
            }


            int len = Math.max(1, goRight - goLeft + 1);
            if(len >= max) {
                max = len;
                range[goLeft] = max;
            }
        }


        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        for(int i = 0; i < n; i++) {
            if(range[i] == max) {
                cnt++;
                if(sb.length() > 0) sb.append(" ");
                sb.append(i+1);
            }
        }
        out.printf("%d %d\n", cnt, max-1);
        out.println(sb.toString());

        out.close();
    }

    static int getRange (int i, int j, int[][] sparse) {
        int gcd = 0;
        int lp = i;
        for(int pow = 19; pow >= 0; pow--) {
            if(lp + (1 << pow) - 1 <= j) {
                gcd = gcd(gcd, sparse[lp][pow]);
                lp += (1 << pow);
            }
        }
        return gcd;
    }

    static int gcd (int a, int b) {
        while (b > 0) {
            int t = a%b;
            a = b;
            b = t;
        }
        return a;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            try	{
                br = new BufferedReader(new InputStreamReader(System.in));
//                br = new BufferedReader(new FileReader("testdata.out"));
                st = new StringTokenizer("");
            } catch (Exception e){e.printStackTrace();}
        }

        public String next() {
            if (st.hasMoreTokens())	return st.nextToken();
            try {st = new StringTokenizer(br.readLine());}
            catch (Exception e) {e.printStackTrace();}
            return st.nextToken();
        }

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

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

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

        public String nextLine() {
            String line = "";
            try {line = br.readLine();}
            catch (Exception e) {e.printStackTrace();}
            return line;
        }

        public Integer[] nextIntegerArray(int n) {
            Integer[] a = new Integer[n];
            for(int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        public int[] nextIntArray(int n) {
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        public char[] nextCharArray() {
            return nextLine().toCharArray();
        }
    }

}