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>
using namespace std;
typedef long long ll;
priority_queue <ll> L;
priority_queue <ll,vector<ll>,greater<ll>> R;
int l[100005],r[100005],n;
ll dl,dr,ans;
int main(){
	scanf("%d",&n);
	scanf("%d%d",&l[1],&r[1]);
	L.push(l[1]),R.push(l[1]);
	for (int i=2;i<=n;i++){
		scanf("%d%d",&l[i],&r[i]);
		int len=r[i]-l[i],len1=r[i-1]-l[i-1];
		dl+=len,dr+=len1;
		ll lmin=L.top()-dl,rmin=R.top()+dr;
		if (l[i]<lmin){
			ans=ans+(lmin-l[i]);
			L.pop(),R.push(lmin-dr);
			L.push(l[i]+dl),L.push(l[i]+dl);
		}else
		if(l[i]>rmin){
			ans=ans+(l[i]-rmin);
			R.pop(),L.push(rmin+dl);
			R.push(l[i]-dr),R.push(l[i]-dr);
		}else{
			L.push(l[i]+dl),R.push(l[i]-dr);
		}
	}
	printf("%lld\n",ans);
} 