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).
#include <bits/stdc++.h>
using namespace std;
int A[400100];
vector<int> v;
int main() {
  int i, j, k, temp, maxx = 0, l, n, sum = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &A[i]);
  for (i = 0; i < n; i++) {
    temp = 0;
    for (k = i - 1; A[i] && A[k] % A[i] == 0 && k >= 0; k--) temp++;
    for (j = i; A[i] && A[j] % A[i] == 0 && j < n; j++) temp++;
    i = j - 1;
    if (temp > maxx) {
      maxx = temp;
      v.clear();
      v.push_back(k + 2);
    } else if (temp == maxx) {
      v.push_back(k + 2);
    }
    sum += temp;
  }
  printf("%d %d\n", v.size(), maxx - 1);
  for (i = 0; i < v.size(); i++) printf("%d ", v[i]);
  return 0;
}
