CodeForces - 359D Pair of Numbers （单调栈）

题目链接：http://codeforces.com/problemset/problem/359/D点击打开链接

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).

#include <iostream>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
#include <limits.h>
#include <math.h>
#include <stack>
#include <queue>
using namespace std;
int gcd(int x,int y)
{
if(x%y==0||y%x==0)
return 1;
return 0;
}
int a[300010];
int la[300010];
int ra[300010];
set<int > q;
set<int > ::iterator it;
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
stack<int > s;
for(int i=1;i<=n;i++)
{
la[i]=i;
if(s.empty())
{
s.push(i);
}
else
{
while(!s.empty())
{
if(a[s.top()]%a[i]==0)
{
la[i]=la[s.top()];
s.pop();
}
else if(a[i]%a[s.top()]==0)
{
s.push(i);
break;
}
else
{
s.push(i);
break;
}
}
if(s.empty())
s.push(i);
}
}
while(!s.empty())
s.pop();

for(int i=n;i>=1;i--)
{
ra[i]=i;
if(s.empty())
{
s.push(i);
}
else
{
while(!s.empty())
{
if(a[s.top()]%a[i]==0)
{
ra[i]=ra[s.top()];
s.pop();
}
else if(a[i]%a[s.top()]==0)
{
s.push(i);
break;
}
else
{
s.push(i);
break;
}
}
if(s.empty())
s.push(i);
}
}

int maxn=0;int cnt=0;

for(int i=1;i<=n;i++)
maxn=max(maxn,ra[i]-la[i]);
for(int i=1;i<=n;)
if((ra[i]-la[i])==maxn)
cnt++,q.insert(la[i]),i=(ra[i]+1);
else
i++;
printf("%d %d\n",cnt,maxn);
for(it=q.begin();it!=q.end();it++)
printf("%d ",(*it));
}