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>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<T> vt;
inline int getInt() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;



int main() {
  int n;
  scanf("%d",&n);
  vi l(n), r(n);
  rep(i,n) cin>>l[i]>>r[i];
  priority_queue<ll> ql;
  PQ(ll) qr;
  ll offl = 0, offr = 0;
  ql.push(0); qr.push(0);
  ll ans = 0;
  rep(i,n-1) {
    ll xl = r[i+1]-l[i];
    ll xr = l[i+1]-r[i];
    offl -= xl; offr -= xr;
    if (offl+ql.top() >= 0) {
      ql.push(-offl); ql.push(-offl);
      ll x = ql.top()+offl; ql.pop();
      ans += x;
      qr.push(x-offr);
    } else if (offr+qr.top() <= 0) {
      qr.push(-offr); qr.push(-offr);
      ll x = qr.top()+offr; qr.pop();
      ans += -x;
      ql.push(x-offl);
    } else {
      ql.push(-offl);
      qr.push(-offr);
    }
  }
  cout<<ans<<endl;
  return 0;
}




















