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
// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=100000+10;

int n,l[N],r[N];
priority_queue<ll> L;
priority_queue<ll,vector<ll>,greater<ll>> R;

int main() {
	n=read();
	for (int i=1;i<=n;++i) l[i]=read(),r[i]=read();
	L.push(l[1]),R.push(l[1]); ll addl=0,addr=0,ans=0;
	for (int i=2;i<=n;++i) {
		addl-=r[i]-l[i],addr+=r[i-1]-l[i-1];
		ll lb=L.top()+addl,rb=R.top()+addr;
		if (lb<=l[i]&&l[i]<=rb) L.push(l[i]-addl),R.push(l[i]-addr);
		else if (l[i]<lb) {
			ans+=lb-l[i];
			R.push(lb-addr);
			L.pop(),L.push(l[i]-addl),L.push(l[i]-addl);
		} else {
			ans+=l[i]-rb;
			L.push(rb-addl);
			R.pop(),R.push(l[i]-addr),R.push(l[i]-addr);
		}
	}
	printf("%lld\n",ans);
	return 0;
}