Problem:
Score : 1000 points

Problem Statement

AtCoDeer the deer found N rectangle lying on the table, each with height 1.
If we consider the surface of the desk as a two-dimensional plane, the i-th rectangle i(1≤i≤N) covers the vertical range of [i-1,i] and the horizontal range of [l_i,r_i], as shown in the following figure:

AtCoDeer will move these rectangles horizontally so that all the rectangles are connected.
For each rectangle, the cost to move it horizontally by a distance of x, is x.
Find the minimum cost to achieve connectivity.
It can be proved that this value is always an integer under the constraints of the problem.

Constraints

All input values are integers.

1≤N≤10^5

1≤l_i<r_i≤10^9

Partial Score

300 points will be awarded for passing the test set satisfying 1≤N≤400 and 1≤l_i<r_i≤400.

Input

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

N
l_1 r_1
l_2 r_2
:
l_N r_N

Output

Print the minimum cost to achieve connectivity.

Sample Input 1

3
1 3
5 7
1 3

Sample Output 1

2

The second rectangle should be moved to the left by a distance of 2.

Sample Input 2

3
2 5
4 6
1 4

Sample Output 2

0

The rectangles are already connected, and thus no move is needed.

Sample Input 3

5
999999999 1000000000
1 2
314 315
500000 500001
999999999 1000000000

Sample Output 3

1999999680

Sample Input 4

5
123456 789012
123 456
12 345678901
123456 789012
1 23

Sample Output 4

246433

Sample Input 5

1
1 400

Sample Output 5

0

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

// 基本テンプレート
// #define _GLIBCXX_DEBUG // for STL debug (optional)
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

#define debug(...) fprintf(stderr, __VA_ARGS__)
#define int long long int
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
typedef pair<int, int> pii;
typedef long long ll;
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const ll INF = 1001001001001001LL;
const ll MOD = 1000000007LL;
 
signed main() {
    int N; cin >> N;
    vector<int> L(N), R(N);
    for(int i=0; i<N; i++) {
        cin >> L[i] >> R[i];
    }

    priority_queue< int > queL;
    priority_queue< int, vector<int>, greater<int> > queR;

    // 最小値の y 座標 (最終的な答え)
    int flr = 0;

    // 最小値より左 / 右にある点は、どれだけシフトしたか？
    int ofsL = 0, ofsR = 0;
    
    queL.push(L[0]);
    queR.push(L[0]);

    for(int i=1; i<N; i++) {
        ofsL += R[i] - L[i];
        ofsR += R[i-1] - L[i-1];

        // 左側領域の右端と、右側領域の左端
        // 区間 [lb, ub] 内の任意の座標について最小値を達成する 
        int lb = queL.top() - ofsL, ub = queR.top() + ofsR;

        // 点を push する際は、その時点での offset をキャンセルするよう注意
        if(lb <= L[i] and L[i] <= ub) {
            // 傾き 0 (最小値を取る領域内) の場合、最小値は変わらない
            queL.push(L[i] + ofsL);
            queR.push(L[i] - ofsR);
        }
        else if(L[i] < lb) {
            // 傾き負の領域内にいた場合、最小値が変化
            flr += lb - L[i];
            // 左側にある点であって元の最小値に最も近い点は、
            // 最小値が変化したことにより傾きが正へ変化する
            queR.push(lb - ofsR);
            queL.pop();
            // L[i] に相当する情報を追加
            queL.push(L[i] + ofsL);
            queL.push(L[i] + ofsL);
        }
        else {
            flr += L[i] - ub;
            queL.push(ub + ofsL);
            queR.pop();
            queR.push(L[i] - ofsR);
            queR.push(L[i] - ofsR);
        }
    }
    cout << flr << endl;
    return 0;
}