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:

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

54 6 9 3 6

output

1 32 

input

51 3 5 7 9

output

1 41 

input

52 3 5 7 11

output

5 01 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 0x3f3f3f3fusing 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;}