// Original C++ solution (slower):
#include <bits/stdc++.h>

#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define PI 3.141592653589793238462643383279502884L

#define lli long long int

#define d double

#define mod 1000000007

#define mx 100001

using namespace std;

lli dp[401][401];//dp[l][r]->min cost to mashup from range i to j....

lli sum[401][401];

lli fun(vector<int>&vec,int l,int r){

	if(l>=r) return 0;

	if(r-l==1) return vec[l]+vec[r];

	if(dp[l][r]!=-1) return dp[l][r];

	lli ans=99999999999999999;

	for(int i=l;i<r;i++){

		ans=min(ans,fun(vec,l,i)+fun(vec,i+1,r)+sum[l][r]);

	}

	return dp[l][r]=ans;

}

int main() {

	FIO;

	int n;

	cin>>n;

	vector<int> vec(n);

	for(int i=0;i<n;i++) cin>>vec[i];

	memset(sum,0,sizeof(sum));

	for(int i=0;i<n;i++){

		for(int j=i;j<n;j++){

			for(int k=i;k<=j;k++) sum[i][j]+=vec[k];

		}

	}

	memset(dp,-1,sizeof(dp));

	lli ans=fun(vec,0,n-1);

	cout<<ans<<endl;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define PI 3.141592653589793238462643383279502884L

#define make_unique(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))

#define Sort(vec) sort(vec.begin(),vec.end())

#define RSort(vec) sort(vec.rbegin(),vec.rend())

#define endl "\n"

#define tr1(i) cout<<i<<endl;

#define tr2(i,j) cout<<i<<" "<<j<<endl;

#define tr3(i,j,k) cout<<i<<" "<<j<<" "<<k<<endl;

#define vvi vector<vector<int> > 

#define vvl vector<vector<ll> >

#define all(st) st.begin(),st.end()

#define vpl vector<pair<ll,ll> >

#define vpi vector<pair<int,int> >

#define pi pair<int,int>

#define pl pair<ll,ll>

#define al vector<list<int> >

#define vs vector<string>

#define vb vector<bool>

#define vi vector<int>

#define vl vector<ll>

#define vc vector<char>

#define rep(i,l,r) for(int i=l;i<r;i++)

#define repit(st) for(auto it=st.begin();it!=st.end();it++)

#define ull unsigned long long int

#define eb emplace_back

#define pb push_back

#define ll long long int

#define minf -(1e18)

#define inf 1e18

#define f first

#define se second

#define mod 1000000007

#define mx 100001

using namespace std;

ll mi(ll n,ll m){ll pw=n%m;ll ex=m-2;ll ans=1;while(ex){if(ex&1) ans = ans*pw%m;pw = pw*pw%m;ex>>=1;}return ans%mod;}

ll pw(ll a,ll n){ll pw=a%mod;ll ex=n;ll ans=1;while(ex){if(ex&1) ans = ans*pw%mod;pw = pw*pw%mod;ex>>=1;}return ans%mod;}

template<typename T>istream &operator>>(istream &is, vector<T> &v) { for (T &t : v) is >> t; return is; }

template<typename T>ostream &operator<<(ostream &os, const vector<T> &v) { for (const T &t : v) {os << t <<" ";} os << '\n'; return os; }

void show(ll x){cout<<x<<" ";}

ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}

ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll max(ll a,ll b){return a>b?a:b;}

ll min(ll a,ll b){return a<b?a:b;}

ll Abs(ll x){return x>0?x:(-x);}

ll Ceil(ll a,ll b){return a/b+(a%b!=0);}

#define bs binary_search

#define lb lower_bound

#define ub upper_bound

#define mkp make_pair

//#define d double

void solve(){

	ll n;

	cin>>n;

	vl a(n);

	cin>>a;

	vvl dp(n,vl(n,inf));

	for(int i=0;i<n;i++){

		dp[i][i]=0;

		if(i<n-1) dp[i][i+1]=a[i]+a[i+1];

	}

	vvl sum(n,vl(n));

	for(int i=0;i<n;i++){

		for(int j=0;j<n;j++){

			for(int k=i;k<=j;k++){

				sum[i][j]+=a[k];

			}

		}

	}

	for(int l=3;l<=n;l++){

		for(int i=0;i<=n-l;i++){

			for(int j=i;j<i+l-1;j++){

				dp[i][i+l-1]=min(dp[i][i+l-1],dp[i][j]+dp[j+1][i+l-1]+sum[i][i+l-1]);

			}

		}

	}

	cout<<dp[0][n-1];

 

 

}

int32_t main(){

int t=1;

//cin>>t;

while(t--){

	solve();

}

}

// Fastest known solution for this problem: