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
/*#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")*/
#include<bits/stdc++.h>
#define ll long long
#define inf 1000000005
#define put putchar('\n')
#define F(i,a,b) for (int i=(a);i<=(b);i++)
#define D(i,a,b) for (int i=(a);i>=(b);i--)
#define go(i,t) for (int i=head[t];i;i=Next[i])
#define sqr(x) ((x)*(x))
#define re register
#define mp make_pair
#define fi first
#define se second
#define pa pair<int,int>
#define pb push_back
#define be begin()
#define en end()
#define ret return puts("-1"),0;
#define mod 1000000007
#define N 500055
#define int ll
using namespace std;
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline int read(){char c=getchar();int tot=1;while ((c<'0'|| c>'9')&&c!='-') c=getchar();if (c=='-'){tot=-1;c=getchar();}
int sum=0;while (c>='0'&&c<='9'){sum=sum*10+c-'0';c=getchar();}return sum*tot;}
inline void wr(int x){if (x<0) {putchar('-');wr(-x);return;}if(x>=10)wr(x/10);putchar(x%10+'0');}
inline void wrn(int x){wr(x);put;}inline void wri(int x){wr(x);putchar(' ');}
inline void wrn(int x,int y){wri(x);wrn(y);}inline void wrn(int a,int b,int c){wri(a);wrn(b,c);}
int n,m,l[N],r[N],addl,addr,ans,l1,l2,lb,rb;
priority_queue <ll> L;
priority_queue<ll, vector<ll>, greater<ll> > R;
signed main(){
	n=read();
	F(i,1,n){
		l[i]=read();r[i]=read();
		if (i==1){L.push(l[i]);R.push(l[i]);continue;}
		l2=r[i]-l[i];l1=r[i-1]-l[i-1];
		addl-=l2;addr+=l1;
		ll lb=L.top()+addl,rb=R.top()+addr;
		if (l[i]<lb){
			ans+=lb-l[i];
			R.push(lb-addr);
			L.push(l[i]-addl);
			L.push(l[i]-addl);L.pop();
		}
		else if (l[i]>rb){
			ans+=l[i]-rb;
			L.push(rb-addl);
			R.push(l[i]-addr);R.push(l[i]-addr);
			R.pop();
		}
		else{
			L.push(l[i]-addl);
			R.push(l[i]-addr);
		}
	}
	wrn(ans);
	return 0;
}