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



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define F first

#define S second

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

#define sz(x) int(x.size())

#define pb push_back



using namespace std;

using ll=long long;

using pii= pair<int, int>;

mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());

template<typename... T> void rd(T&... args) {((cin>>args), ...);}

template<typename... T> void wr(T... args) {((cout<<args<<" "), ...);cout<<endl;}



const int N=405;

ll dp[N][N];

int main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

    int n;

    cin>>n;

    vector<ll> sum(n+1),a(n);

    memset(dp,0x3f,sizeof(dp));

    forn(i,n){

        cin>>a[i];

        dp[i][i]=0;

    }

    partial_sum(all(a),sum.begin()+1);

    for(int len=2;len<=n;len++){

        for(int l=0;l+len-1<n;l++){

            for(int k=l;k<l+len-1;k++){

                dp[l][l+len-1]=min(dp[l][l+len-1],dp[l][k]+dp[k+1][l+len-1]+sum[l+len]-sum[l]);

            }

        }

    }

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

    return 0;

}




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

#define f first

#define s second

#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)

#define pb push_back

#define all(s) begin(s), end(s)

#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define sz(s) int(s.size())

#define ENDL '\n'

using namespace std;

typedef long double ld;

typedef long long lli;

typedef pair<lli,lli> ii;

typedef vector<int> vi;

typedef vector<lli> vl;

typedef vector<ii> vii;

typedef vector<vi> vvi;

typedef vector<vl> vvl;

#define deb(x) cout << #x": " << (x) << endl;



const int N = 3e3+5;

const lli INF = 1e18;

const lli MOD = 1e9+7;



vi m;

vi primes(1,2);

vector<vector<ld>>nk;

void sieve(){

	m = vi(N+1,0);

	for(int i = 4; i<N; i+=2)m[i]=1;

	for(int i = 3; i*i<=N; i+=2)if(!m[i])for(int j = i*i; j<N; j+=i)m[j]=1;

	for(int i = 3; i<N; i+=2)if(!m[i])primes.pb(i);

}



void pascal(){

	nk =vector<vector<ld>>(N,vector<ld>(N,0.0));

	fore(i,0,N)nk[i][0]=nk[i][i]=1.0;

	fore(i,1,N)fore(j,1,i)nk[i][j]=nk[i-1][j-1]+nk[i-1][j];

}



lli gcd(lli a, lli b){return (b?gcd(b,a%b):a);}



lli lcm(lli a, lli b){

	if(a<b)swap(a,b);

	lli c = gcd(a,b);

	a/=c;

	return a*b;

}



int popcount(lli x) { return __builtin_popcountll(x); }



lli poww(lli a, lli b){

	lli res =1;

	while(b){

		if(b&1) res = res * a;

		a = a*a;

		b/=2;

	}

	return res;

}



lli powm(lli a, lli b){

	lli res =1;

	while(b){

		if(b&1) res = (res * a)%MOD;

		a = (a*a)%MOD;

		b/=2;

	}

	return res;

}



bool isPrime(lli x){

	if(x==1)return 0;

	for(auto i:primes){

		if(i*i>x)return 1;

		if(x%i==0)return 0;

	}

	return 1;

}



lli SUM(lli a, lli b, lli mod=MOD){

	return (a+b+mod)%mod;

}



int dx[4]={-1,0,0,1}, dy[4]={0,1,-1,0};



// ---- コーディングはここから！ ('-')7



void solve(){

	int n; cin>>n;

	vl v(n+1,0); fore(i,0,n)cin>>v[i];

	vvl dp(n,vl(n,-1));

	vl sum(n+1);

	sum[0]=v[0];

	fore(i,1,n+1)sum[i]=sum[i-1]+v[i];

	auto f=[&](auto f, int l, int r)->lli{

		if(l==r)return 0;

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

		dp[l][r]=INF;

		fore(i,l,r)dp[l][r]=min(dp[l][r],f(f,l,i)+f(f,i+1,r)+sum[r]-(l?sum[l-1]:0));

		return dp[l][r];

	};

	cout<<f(f,0,n-1)<<ENDL;

}



int main(){_

	//int t; cin>>t; while(t--)

	solve();

}


// Fastest known solution for this problem: