From top to bottom, each fence has only left and right endpoints, and from each point only stops at one fence, so each point has two sides at most. We can use the segment tree to maintain this connection. (Each point is connected only when it is covered for the first time, and then two endpoints are added in each time.) Then it is connected again from point s, and the last one is connected to the end point, and then the shortest distance can be run.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <deque>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define N 50010
#define M 200010
#define treen 200001
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,s,num=0,h[N<<1],id1,id2,tot=1;
ll d[N<<1];
bool inq[N<<1];
struct node{
    int x,id;
}tree[M<<2];
struct dat{
    int x,y;
}a[N];
struct edge{
    int to,next,val;
}data[N<<2];
inline void add(int x,int y,int val){
    data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].val=val;
}
void build(int p,int l,int r){
    if(l==r) return;
    int mid=l+r>>1;
    build(p<<1,l,mid);build(p<<1|1,mid+1,r);
}
void solve(int p,int l,int r,int x,int y){
    if(tree[p].x==0) return;
    if(l==r){
        if(id2==id1) add(tree[p].id,id1,abs(l-100001));
        else add(tree[p].id,id1,l-x),add(tree[p].id,id2,y-l);
        tree[p].x=0;return;
    }
    int mid=l+r>>1;
    if(x<=mid) solve(p<<1,l,mid,x,y);
    if(y>mid) solve(p<<1|1,mid+1,r,x,y);
    tree[p].x=tree[p<<1].x+tree[p<<1|1].x;
}
void insert1(int p,int l,int r,int x,int val){
    if(l==r){tree[p].x=1,tree[p].id=val;return;}
    int mid=l+r>>1;
    if(x<=mid) insert1(p<<1,l,mid,x,val);
    else insert1(p<<1|1,mid+1,r,x,val);
    tree[p].x=tree[p<<1].x+tree[p<<1|1].x;
}
void spfa(){
    deque<int>q;
    for(int i=1;i<=tot;++i) d[i]=1LL<<60;d[1]=0;
    q.push_back(1);inq[1]=1;
    while(!q.empty()){
        int x=q.front();q.pop_front();inq[x]=0;
        for(int i=h[x];i;i=data[i].next){
            int y=data[i].to;
            if(d[x]+data[i].val<d[y]){
                d[y]=d[x]+data[i].val;
                if(!inq[y]){
                    if(!q.empty()&&d[y]<d[q.front()]) q.push_front(y);
                    else q.push_back(y);inq[y]=1;
                }
            }
        }
    }
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();s=read()+100001;
    build(1,1,treen);
    for(int i=1;i<=n;++i){
        a[i].x=read()+100001,a[i].y=read()+100001;
    }add(1,2,s-a[n].x);add(1,3,a[n].y-s);
    for(int i=n;i>=1;--i){
        id1=++tot;id2=++tot;
        solve(1,1,treen,a[i].x,a[i].y);
        insert1(1,1,treen,a[i].x,id1);
        insert1(1,1,treen,a[i].y,id2);
    }id1=id2=++tot;solve(1,1,treen,1,treen);
    spfa();
    printf("%lld\n",d[tot]);
    return 0;
}