#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int MX=2e5+5;
int N;
ll A[MX];
ld f(int l, int r) {
return (ld)(A[r]-A[l-1])/(r-l+1);
}
vector<pair<ld,int>> v;
ld L[MX], R[MX];
int idL[MX], idR[MX];
struct fenwick {
int t[2*MX];
void upd(int pos, int val) {
for(int i=pos;i<=2*N;i+=i&-i) t[i]+=val;
}
int que(int pos) {
int res=0;
for(int i=pos;i>0;i-=i&-i) res+=t[i];
return res;
}
int que(int l, int r) {
return que(r)-que(l-1);
}
} fw;
int main() {
cin.tie(0); ios_base::sync_with_stdio(0);
cin>>N;
for(int i=1;i<=N;i++) {
cin>>A[i];
A[i]+=A[i-1];
}
for(int k=1;k<=N;k++) {
int l=k,r=N,res=0;
while(l<=r) {
int mid=(l+r)/2;
if(f(mid-k+1,mid)>=f(1,N)) {
res=mid,r=mid-1;
} else {
l=mid+1;
}
}
// f(res-k,res-1) or f(res-k+1,res)
if(res-k<1)
L[k]=f(res-k+1,res);
else
L[k]=f(res-k,res-1);
R[k]=f(res-k+1,res);
v.push_back({L[k],k});
v.push_back({R[k],k});
}
sort(v.begin(),v.end());
int id=1;
for(auto [x,k]:v) {
if(!idL[k]) idL[k]=id;
else idR[k]=id;
id++;
}
for(int k=1;k<=N;k++) fw.upd(idL[k],1);
/*
each k adds 2 important point f(l,l+k-1) or f(l+1,l+k)
for some l s.t. f(l,l+k-1) <= f(1,N) <= f(l+1,l+k)
a range [L,R] is valid iff every k in [1,N] exist atleast one important point
we can show sufficiency by noticing that
(*) f(l,r) < f(l+1,r+1) < f(l+2,r+2) < ...
also notice that important points are two consecutive terms in (*)
now plot each f(l,r) in a similar fashion to pascal's triangle
where each row groups all ranges with same length
assume we reach the k-th row and want to go down to (k+1) th row
if we're in the left imp. point of k-th row which is in [L,R]
the option is to either go to the left imp. point of (k+1)-th row of right imp. point
if left imp. point is in [L,R] we are done
if left imp. point isn't in [L,R] then it must either be in [1,L) or (R,N]
if it is in (R,N], this is absurd due to our assumption that there exist an imp point of range
(k+1) in [L,R]
if it is in [1,L), then the right imp. point must be in [L,R] so we can go there and we are done.
this is because since we are in left imp. point of k-th row
it can produce to imp. point in k+1 th row
also notice that f(l,r-1) < f(l,r) < f(l+1,r)
so what it produces is one is to its left and one is to its right
since one is already to its left, the other one must be to its right
*/
ld ans=2e18;
for(auto [x,k]:v) {
int l=1,r=2*N,res=0;
while(l<=r) {
int mid=(l+r)/2;
if(fw.que(mid)==N) {
r=mid-1,res=mid;
} else {
l=mid+1;
}
}
if(res!=0) {
ans=min(ans,v[res-1].first-x);
}
if(idL[k]) {
fw.upd(idL[k],-1);
fw.upd(idR[k],1);
idL[k]=0;
} else {
fw.upd(idR[k],-1);
}
}
cout<<fixed<<setprecision(10)<<ans<<'\n';
}
