#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define l first
#define r second
const int N=3e3+5;
int a[N];
ll pre[N];
ll dp[N][N],mn[N];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i],pre[i]=pre[i-1]+a[i];
for(int i=0;i<=n;i++)
{
for(int j=0;j<=n;j++)
{
dp[i][j]=-1e9;
}
}
dp[0][0]=0;
ll ans=0;
for(int i=1;i<=n;i++)
{
dp[i][i]=1;
mn[i]=1e9;
// cout<<"If we fix "<<i<<endl;
for(int j=1;j<=i;j++)
{
// k is a range
// as k increases pre[i-j-k] decreases
// find minimum k that satisfy the condition
int s=-1,e=i-j+1;
while(s+1<e)
{
int mid=(s+e)/2;
if(pre[i-j-mid]>=2*pre[i-j]-pre[i])
{
e=mid;
}
else
{
s=mid;
}
}
if(e!=(i-j+1))
{
dp[i][j]=max(dp[i][j],dp[i-j][e]+1);
}
else
{
// no k this dp is null
}
e=mn[i-j];
if(e!=1e9 and pre[i-j-e]>=2*pre[i-j]-pre[i])
{
dp[i][j]=max(dp[i][j],dp[i-j][e]+1);
}
if(dp[i][j]>0 and mn[i]==1e9)
{
mn[i]=j;
}
ans=max(ans,dp[i][j]);
}
}
cout<<ans<<endl;
}
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