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;
#define N 120000

LL n,sl,sr,ans,l[N],r[N],len[N];
priority_queue<LL> L;
priority_queue<LL,vector<LL>,greater<LL> > R;

int main(){
	scanf("%lld",&n);
	for (LL i=1;i<=n;++i){scanf("%lld%lld",l+i,r+i); len[i]=r[i]-l[i];}
	L.push(r[1]); R.push(r[1]);
	for (LL i=2;i<=n;++i){
		sl-=len[i-1]; sr+=len[i];
		LL lp=L.top()+sl,rp=R.top()+sr;
		if (r[i]>=lp&&r[i]<=rp){
			L.push(r[i]-sl); R.push(r[i]-sr);
		}
		if (r[i]>rp){
			ans+=r[i]-rp; R.pop(); L.push(rp-sl);
			R.push(r[i]-sr); R.push(r[i]-sr);
		}
		if (r[i]<lp){
			ans+=lp-r[i]; L.pop(); R.push(lp-sr);
			L.push(r[i]-sl); L.push(r[i]-sl);
		}
	}
	printf("%lld\n",ans);
	
	return 0;
}
			