/*
类似于网络流24题中的一道（求最长上升子序列的最多个数）：
构图方案：
首先求出LIS；
然后拆点 建边Edge(i,i+n,1) 因为没个点只能用一次，所以边的容量为1
然后对于每个dp[i]==1的建边Edge(s,i,1)
dp[i]==k的建边Edge(i+n,t,1);
dp[i]==dp[j]+1 建边Edge(j+n,i,1);
然后sap 求最大流就OK了
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define M 10100
int gap[M],dis[M],pre[M],cur[M],dp[M];
int NE,NV,sink,source;
int head[M],p[M],k,n;
struct Node
{
int c,pos,next;
} E[999999];
#define FF(i,NV) for(int i=0;i<NV;i++)
int sap(int s,int t)
{
//memset(pre,-1,sizeof(pre));
memset(dis,0,sizeof(int)*(NV+1));
memset(gap,0,sizeof(int)*(NV+1));
FF(i,NV) cur[i] = head[i];
int u = pre[s] = s,maxflow = 0,aug = 1<<29;
gap[0] = NV;
while(dis[s] < NV)
{
loop:
for(int &i = cur[u]; i != -1; i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && dis[u] == dis[v] + 1)
{
aug=min(aug,E[i].c);
pre[v] = u;
u = v;
if(v == t)
{
maxflow += aug;
for(u = pre[u]; v != s; v = u,u = pre[u])
{
E[cur[u]].c -= aug;
E[cur[u]^1].c += aug;
}
aug = 1<<29;
}
goto loop;
}
}
int mindis = NV;
for(int i = head[u]; i != -1 ; i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && mindis > dis[v])
{
cur[u] = i;
mindis = dis[v];
}
}
if( (--gap[dis[u]]) == 0)break;
gap[ dis[u] = mindis+1 ] ++;
u = pre[u];
}
return maxflow;
}
void addEdge(int u,int v,int c )
{
E[NE].c = c;
E[NE].pos = v;
E[NE].next = head[u];
head[u] = NE++;
E[NE].c = 0;
E[NE].pos = u;
E[NE].next = head[v];
head[v] = NE++;
}
int LIS()
{
int i,j,ans=0;
dp[1]=1;
for(i=2;i<=n;++i)
{
dp[i]=1;
for(j=1;j<i;++j)
{
if(p[j]<p[i]&&dp[j]>=dp[i])dp[i]=dp[j]+1;
}
ans=max(ans,dp[i]);
}
return ans;
}
int main()
{
int i,j;
while(scanf("%d",&n)!=EOF)
{
NE=source=0;
sink=n<<1|1;
NV=sink+1;
memset(head,-1,sizeof(head));
// memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
scanf("%d",&p[i]);
}
k=LIS();
for(i=1;i<=n;i++)
{
addEdge(i,i+n,1);
if(dp[i]==1)
addEdge(source,i,1);
if(dp[i]==k)
addEdge(i+n,sink,1);
for(j=i+1;j<=n;++j)
if(dp[j]==dp[i]+1)addEdge(i+n,j,n);
}
printf("%d\n%d\n",k,sap(source,sink));
}
return 0;
}
/*
4
3 6 2 5
*/
