Problem: Annie has gotten bored of winning every coding contest and farming unlimited rating. Today, she is going to farm potatoes instead.

Annie's garden is an infinite 2D plane. She has n potatoes to plant, and the i-th potato must be planted at (x_i,y_i). Starting at the point (0, 0), Annie begins walking, in one step she can travel one unit right or up (increasing her x or y coordinate by 1 respectively). At any point (X,Y) during her walk she can plant some potatoes at arbitrary points using her potato gun, consuming max(|X-x|,|Y-y|) units of energy in order to plant a potato at (x,y). Find the minimum total energy required to plant every potato.

Note that Annie may plant any number of potatoes from any point.

Input

The first line contains the integer n (1 ≤ n ≤ 800 000).

The next n lines contain two integers x_i and y_i (0 ≤ x_i,y_i ≤ 10^9), representing the location of the i-th potato. It is possible that some potatoes should be planted in the same location.

Output

Print the minimum total energy to plant all potatoes.

Examples

Input


2
1 1
2 2


Output


0


Input


2
1 1
2 0


Output


1


Input


3
5 5
7 7
4 9


Output


2


Input


10
5 1
4 0
9 6
0 2
10 1
9 10
3 10
0 10
8 9
1 5


Output


19


Input


10
1 1
2 2
2 0
4 2
4 0
2 0
0 2
4 0
4 2
5 1


Output


6

Note

In example 1, Annie can travel to each spot directly and plant a potato with no energy required.

In example 2, moving to (1,0), Annie plants the second potato using 1 energy. Next, she travels to (1,1) and plants the first potato with 0 energy.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std ;

#define ll              long long 
#define pb              push_back
#define all(v)          v.begin(),v.end()
#define sz(a)           (ll)a.size()
#define F               first
#define S               second
#define INF             2000000000000000000
#define popcount(x)     __builtin_popcountll(x)
#define pll             pair<ll,ll>
#define pii             pair<int,int>
#define ld              long double

const int M = 1000000007;
const int MM = 998244353;

template<typename T, typename U> static inline void amin(T &x, U y){ if(y<x) x=y; }
template<typename T, typename U> static inline void amax(T &x, U y){ if(x<y) x=y; }

#ifdef LOCAL
#define debug(...) debug_out(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) 2351
#endif


int _runtimeTerror_()
{
    int n;
    cin>>n;
    vector<pii> a;
    for(int i=0;i<n;++i)
    {
        int x,y;
        cin>>x>>y;
        a.push_back({x+y,x-y});
    }
    sort(a.begin(),a.end());
    // debug(a);
    priority_queue<int,vector<int>,greater<int>> R;
    priority_queue<int> L;
    L.push(0);
    R.push(0);
    int64_t ans = 0;
    int pr = 0;
    int l_sh = 0,r_sh = 0;
    for(auto &[x,y]:a)
    {
        int d = abs(x - pr);
        pr = x;
        l_sh -= d, r_sh += d;
        // debug(L.top()+l_sh,R.top()+r_sh,y);
        if(y >= L.top() + l_sh && y <= R.top() + r_sh)
        {
            L.push(y-l_sh);
            R.push(y-r_sh);
        }
        else if(y > R.top() + r_sh)
        {
            ans += abs(y-R.top()-r_sh);
            R.push(y-r_sh);
            R.push(y-r_sh);
            L.push(R.top()+r_sh-l_sh);
            R.pop();
        }
        else
        {
            ans += abs(y - L.top()-l_sh);
            L.push(y-l_sh);
            L.push(y-l_sh);
            R.push(L.top()+l_sh-r_sh);
            L.pop();
        }
    }
    ans /= 2;
    cout << ans << "\n";
    return 0;
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    #ifdef runSieve
        sieve();
    #endif
    #ifdef NCR
        initialize();
    #endif
    int TESTS=1;
    //cin>>TESTS;
    while(TESTS--)
        _runtimeTerror_();
    return 0;
}