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:
/*
    author:  Maksim1744
    created: 14.06.2021 03:12:24
*/

#include "bits/stdc++.h"

using namespace std;

using ll = long long;
using ld = long double;

#define mp   make_pair
#define pb   push_back
#define eb   emplace_back

#define sum(a)     ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)    (*min_element((a).begin(), (a).end()))
#define maxe(a)    (*max_element((a).begin(), (a).end()))
#define mini(a)    ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)    ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())

template<typename T>             vector<T>& operator--            (vector<T> &v){for (auto& i : v) --i;            return  v;}
template<typename T>             vector<T>& operator++            (vector<T> &v){for (auto& i : v) ++i;            return  v;}
template<typename T>             istream& operator>>(istream& is,  vector<T> &v){for (auto& i : v) is >> i;        return is;}
template<typename T>             ostream& operator<<(ostream& os,  vector<T>  v){for (auto& i : v) os << i << ' '; return os;}
template<typename T, typename U> pair<T,U>& operator--           (pair<T, U> &p){--p.first; --p.second;            return  p;}
template<typename T, typename U> pair<T,U>& operator++           (pair<T, U> &p){++p.first; ++p.second;            return  p;}
template<typename T, typename U> istream& operator>>(istream& is, pair<T, U> &p){is >> p.first >> p.second;        return is;}
template<typename T, typename U> ostream& operator<<(ostream& os, pair<T, U>  p){os << p.first << ' ' << p.second; return os;}
template<typename T, typename U> pair<T,U> operator-(pair<T,U> a, pair<T,U> b){return mp(a.first-b.first, a.second-b.second);}
template<typename T, typename U> pair<T,U> operator+(pair<T,U> a, pair<T,U> b){return mp(a.first+b.first, a.second+b.second);}
template<typename T, typename U> void umin(T& a, U b){if (a > b) a = b;}
template<typename T, typename U> void umax(T& a, U b){if (a < b) a = b;}

#ifdef HOME
#define SHOW_COLORS
#include "C:/C++ libs/print.cpp"
#else
#define show(...) void(0)
#define mclock    void(0)
#define shows     void(0)
#define debug  if (false)
#endif

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    int n;
    cin >> n;
    vector<pair<int, int>> v(n);
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        v[i] = {x + y, y};
    }
    sort(v.begin(), v.end());

    multiset<int> left = {0}, right = {0};
    int right_shift = 0;

    ll total = 0;
    int lastx = 0;

    for (auto [x, y] : v) {
        total += y;
        right_shift += x - lastx;

        int lend = *left.rbegin();
        int rbegin = *right.begin() + right_shift;
        assert(lend <= rbegin);
        if (y < lend) {
            right.insert(lend - right_shift);
            left.insert(y);
            left.insert(y);
            left.erase(--left.end());
        } else if (y <= rbegin) {
            left.insert(y);
            right.insert(y - right_shift);
        } else {
            left.insert(rbegin);
            right.insert(y - right_shift);
            right.insert(y - right_shift);
            right.erase(right.begin());
        }
        lastx = x;
    }

    int last = 0;
    while (!left.empty()) {
        int x = *left.begin();
        total -= (ll)(x - last) * left.size();
        left.erase(left.begin());
        last = x;
    }
    cout << total << '\n';

    return 0;
}