#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=100010;
int num[maxn],n,sum[maxn],fmin[maxn][20],fmax[maxn][20];
long long ans=0,ansl,ansr;int querymin(int l,int r)
{int k=0;if (l>r)return 0;while (1<<(k+1)<=r-l+1)k++;return min(fmin[l][k],fmin[r-(1<<k)+1][k]);
}int querymax(int l,int r)
{int k=0,tmp=0;if (l>r)return 0;while (1<<(k+1)<=r-l+1)k++;tmp=max(fmax[l][k],fmax[r-(1<<k)+1][k]);tmp=max(tmp,(sum[r]-sum[l-1])*querymin(l,r));if (tmp>ans){ans=tmp;ansl=l;ansr=r;}return tmp;
}void findmin()                                          //用RMQ求第i个到第j个这个区间的最小值是哪一个
{int i,j;memset(fmin,0x7f,sizeof(fmin));for (i=1;i<=n;i++)fmin[i][0]=i;for (j=1;(1<<j)<=n;j++)for (i=1;i+(1<<j)-1<=n;i++)if (num[fmin[i][j-1]]<num[fmin[i+(1<<(j-1))][j-1]])fmin[i][j]=fmin[i][j-1];else fmin[i][j]=fmin[i+(1<<(j-1))][j-1];
}void findmax()                                           //表示求第i个到第j个这个区间能取得的最大值
{int i,j,l,r;memset(fmax,0,sizeof(fmax));for (i=1;i<=n;i++)fmax[i][0]=num[i]*num[i];for (j=1;(1<<j)<=n;j++)for (i=1;i+(1<<j)-1<=n;i++){r=i+(1<<j)-1;l=querymin(i,r);fmax[i][j]=max(querymax(i,max(i,l-1)),querymax(max(l+1,r),r));fmax[i][j]=max(fmax[i][j],(sum[r]-sum[i-1])*num[l]);if (fmax[i][j]>ans){ans=fmax[i][j];ansl=i;ansr=r;}}
}int main()
{int i,j;scanf("%d",&n);sum[0]=0;for (i=1;i<=n;i++){scanf("%d",&num[i]);sum[i]=sum[i-1]+num[i];}findmin();findmax();cout<<ans<<endl<<ansl<<" "<<ansr<<endl;return 0;
}