Problem:
Score : 1000 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

Snuke received N intervals as a birthday present.
The i-th interval was [-L_i, R_i].
It is guaranteed that both L_i and R_i are positive.
In other words, the origin is strictly inside each interval.

Snuke doesn't like overlapping intervals, so he decided to move some intervals.
For any positive integer d, if he pays d dollars, he can choose one of the intervals and move it by the distance of d.
That is, if the chosen segment is [a, b], he can change it to either [a+d, b+d] or [a-d, b-d].

He can repeat this type of operation arbitrary number of times.
After the operations, the intervals must be pairwise disjoint (however, they may touch at a point).
Formally, for any two intervals, the length of the intersection must be zero.

Compute the minimum cost required to achieve his goal.

Constraints

1 ≤ N ≤ 5000

1 ≤ L_i, R_i ≤ 10^9

All values in the input are integers.

Input

The input is given from Standard Input in the following format:

N
L_1 R_1
:
L_N R_N

Output

Print the minimum cost required to achieve his goal.

Sample Input 1

4
2 7
2 5
4 1
7 5

Sample Output 1

22

One optimal solution is as follows:

Move the interval [-2, 7] to [6, 15] with 8 dollars

Move the interval [-2, 5] to [-1, 6] with 1 dollars

Move the interval [-4, 1] to [-6, -1] with 2 dollars

Move the interval [-7, 5] to [-18, -6] with 11 dollars

The total cost is 8 + 1 + 2 + 11 = 22 dollars.

Sample Input 2

20
97 2
75 25
82 84
17 56
32 2
28 37
57 39
18 11
79 6
40 68
68 16
40 63
93 49
91 10
55 68
31 80
57 18
34 28
76 55
21 80

Sample Output 2

7337

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// {{{ by shik
#include <bits/stdc++.h>
#include <unistd.h>
#define SZ(x) ((int)(x).size())
#define ALL(x) begin(x),end(x)
#define REP(i,n) for ( int i=0; i<int(n); i++ )
#define REP1(i,a,b) for ( int i=(a); i<=int(b); i++ )
#define FOR(it,c) for ( auto it=(c).begin(); it!=(c).end(); it++ )
#define MP make_pair
#define PB push_back
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef vector<int> VI;

#ifdef SHIK
template<typename T>
void _dump( const char* s, T&& head ) { cerr<<s<<"="<<head<<endl; }

template<typename T, typename... Args>
void _dump( const char* s, T&& head, Args&&... tail ) {
    int c=0;
    while ( *s!=',' || c!=0 ) {
        if ( *s=='(' || *s=='[' || *s=='{' ) c++;
        if ( *s==')' || *s==']' || *s=='}' ) c--;
        cerr<<*s++;
    }
    cerr<<"="<<head<<", ";
    _dump(s+1,tail...);
}

#define dump(...) do { \
    fprintf(stderr, "%s:%d - ", __PRETTY_FUNCTION__, __LINE__); \
    _dump(#__VA_ARGS__, __VA_ARGS__); \
} while (0)

template<typename Iter>
ostream& _out( ostream &s, Iter b, Iter e ) {
    s<<"[";
    for ( auto it=b; it!=e; it++ ) s<<(it==b?"":" ")<<*it;
    s<<"]";
    return s;
}

template<typename A, typename B>
ostream& operator <<( ostream &s, const pair<A,B> &p ) { return s<<"("<<p.first<<","<<p.second<<")"; }
template<typename T>
ostream& operator <<( ostream &s, const vector<T> &c ) { return _out(s,ALL(c)); }
template<typename T, size_t N>
ostream& operator <<( ostream &s, const array<T,N> &c ) { return _out(s,ALL(c)); }
template<typename T>
ostream& operator <<( ostream &s, const set<T> &c ) { return _out(s,ALL(c)); }
template<typename A, typename B>
ostream& operator <<( ostream &s, const map<A,B> &c ) { return _out(s,ALL(c)); }
#else
#define dump(...)
#endif

template<typename T>
void _R( T &x ) { cin>>x; }
void _R( int &x ) { scanf("%d",&x); }
void _R( long long &x ) { scanf("%" PRId64,&x); }
void _R( double &x ) { scanf("%lf",&x); }
void _R( char &x ) { scanf(" %c",&x); }
void _R( char *x ) { scanf("%s",x); }

void R() {}
template<typename T, typename... U>
void R( T& head, U&... tail ) {
    _R(head);
    R(tail...);
}

template<typename T>
void _W( const T &x ) { cout<<x; }
void _W( const int &x ) { printf("%d",x); }
template<typename T>
void _W( const vector<T> &x ) {
    for ( auto i=x.cbegin(); i!=x.cend(); i++ ) {
        if ( i!=x.cbegin() ) putchar(' ');
        _W(*i);
    }
}

void W() {}
template<typename T, typename... U>
void W( const T& head, const U&... tail ) {
    _W(head);
    putchar(sizeof...(tail)?' ':'\n');
    W(tail...);
}

#ifdef SHIK
#define FILEIO(...)
#else
#define FILEIO(name) do {\
    freopen(name ".in","r",stdin); \
    freopen(name ".out","w",stdout); \
} while (0)
#endif

// }}}

const int N=5140;
const LL INF=9e18;

inline void chkmin( LL &a, LL b ) {
    if ( b<a ) a=b;
}

struct P {
    LL l,r;
} p[N];

bool operator<( P a, P b ) {
    return a.r+a.l>b.r+b.l;
}

int n,w;
LL dp[N],ndp[N];

LL solve( int off ) {
    REP(i,n) {
        p[i].l-=off;
        p[i].r+=off;
    }
    memset(dp,0,sizeof(dp));
    REP(i,n) {
        REP(j,N) ndp[j]=INF;
        REP1(j,0,i) {
            chkmin(ndp[j+1],dp[j]+p[i].l+(p[i].r+p[i].l)*j);
            chkmin(ndp[j],dp[j]+p[i].r+(p[i].r+p[i].l)*(i-j));
        }
        memcpy(dp,ndp,sizeof(dp));
    }
    LL ans=INF;
    REP1(i,0,n) if ( dp[i]<ans ) {
        ans=dp[i];
        w=i;
    }
    REP(i,n) {
        p[i].l+=off;
        p[i].r-=off;
    }
    return ans;
}

int main() {
    R(n);
    REP(i,n) R(p[i].l,p[i].r);
    if ( n==1 ) {
        W(0);
        return 0;
    }
    sort(p,p+n);
    LL ans=INF;
    ans=min(ans,solve(0));
    LL ml=1e18,mr=1e18;
    REP(i,n) {
        ml=min(ml,p[i].l);
        mr=min(mr,p[i].r);
    }
    LL l=-ml,r=mr;
    while ( l!=r ) {
        int m=l+(r-l)/2;
        ans=min(ans,solve(m));
        if ( w<n/2 ) l=m+1;
        else r=m;
    }
    W(ans);
    return 0;
}