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:

<image>

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

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.

Examples

Input

3
1 3
5 7
1 3


Output

2


Input

3
2 5
4 6
1 4


Output

0


Input

5
999999999 1000000000
1 2
314 315
500000 500001
999999999 1000000000


Output

1999999680


Input

5
123456 789012
123 456
12 345678901
123456 789012
1 23


Output

246433


Input

1
1 400


Output

0
#include <bits/stdc++.h>
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
using namespace std;
#define X first
#define Y second
#define pb push_back
typedef pair<int, int> ii;
typedef long long ll;

priority_queue< ll > lower, higher;
ll loz, hiz;

ll opt;

int main()
{
	int n; scanf("%d", &n);
	int xp = 0, yp = 0;
	for(int i = 1; i<= n; i++)
	{
		int x, y; scanf("%d %d", &x, &y);
		loz -= y-x;
		hiz += yp-xp;
		xp = x; yp = y;
		if(!lower.empty() && x<= lower.top()+loz)
		{
			ll k = lower.top()+loz;
			lower.pop();
			opt += abs(x-k);
			higher.push(-(k-hiz));
			lower.push(x-loz);
			lower.push(x-loz);
		}
		else if(!higher.empty() && x>= -higher.top()+hiz)
		{
			ll k = -higher.top()+hiz;
			higher.pop();
			opt += abs(x-k);
			lower.push(k-loz);
			higher.push(-(x-hiz));
			higher.push(-(x-hiz));
		}
		else
		{
			lower.push(x-loz);
			higher.push(-(x-hiz));
		}
	}
	printf("%lld\n", opt);
}