CF359D：Pair of Numbers（数论）

reference：WannaflyUnion Wechat
D. Pair of Numbers

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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:

there is integer j (l ≤ j ≤ r),
such that all integers al, al + 1, ..., ar are
divisible by aj;

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).

题意：给定一个数组，求最长的区间，使得该区间内存在一个元素，它能整除该区间内每一个元素。

思路：从数组第一个元素开始向两边延伸，如右边延伸到r，下一轮从r+1开始考虑即可。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <cmath>
# include <vector>
# include <algorithm>
# define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 3e5;
int a[MAXN+3]={0};
int main()
{
vector<int>v;
int n, ans;
while(~scanf("%d",&n))
{
bool flag = false;
ans = -INF;
for(int i=1; i<=n; ++i)
scanf("%d",&a[i]);
for(int i=1; i<=n; ++i)
{
if(a[i]==1)
{
flag = true;
printf("1 %d\n1\n",n-1);
break;
}
int j, k;
for(j=i; j-1>0&&a[j-1]%a[i]==0; --j);
for(k=i; k+1<=n&&a[k+1]%a[i]==0; ++k);
if(k-j > ans)
{
ans = k-j;
v.clear();
}
if(k-j == ans)
v.push_back(j);
i = k;
}
if(!flag)
{
printf("%d %d\n",v.size(), ans);
for(int i=0; i<v.size()-1; ++i)
printf("%d ",v[i]);
printf("%d\n",v[v.size()-1]);
}
}
return 0;
}