Pair of Numbers

Pair of Numbers

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
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define PI acos(-1)
#define INF 0x3f3f3f3f

int n;
int a[1000006];
int num;
int main(){
	cin>>n;
	for(int i=0;i v;
	
	int maxlen=0;
	for(int i=0;i=0;k--){
			if(a[k]%a[i]==0){
				lenR++; 
			}
			else break;
		}
		
		if(maxlen==lenL+lenR){
			v.push_back(i-lenR+1);
			//num++; 
		}
		else if ( maxlen<(lenL+lenR) ){
			v.clear();
			v.push_back(i-lenR+1);
			maxlen= lenL+lenR ;
			//num=1;
		}	
		i=i+lenL;
	}
	printf("%d %d\n",v.size(),maxlen);
	sort(v.begin(),v.end());
	for(int i=0;i

 

思路很简单,就是枚举。从左到右开始不断地枚举。但是如果只是这样的话,肯定会超时的。于是就想着“剪枝” 把没有必要再遍历的就直接跳过。通过发现,公因数是几个数里面最小的那个。从左到右遍历,每次都以枚举点为中心向两边取余,结束后(不能整除时)把i直接跳到取余时所到达的最右边。即 i=i+lenL; 这行代码最为重要。节省了大量的时间

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