#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
typedef long long ll;
const int N_MAX = 100000+5;

struct Interval {
    int Left, Right;
    Interval(int Left=0,int Right=0):Left(Left),Right(Right) {}
    bool operator <(const Interval&b) const{
        if (this->Left != b.Left)return this->Left < b.Left;
        else return this->Right < b.Right;
    }
}interval[N_MAX];
vector<Interval>vec;
int n, m;
ll c[N_MAX];//c[j]记录到区间j为止(包括j)的所有区间和，不包含中间的空隙
int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        vec.clear();
        memset(c, 0, sizeof(c));
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &interval[i].Left, &interval[i].Right);
        }
        sort(interval, interval + n);
        vec.push_back(interval[0]);
        for (int i = 1; i < n; i++) {//合并所有可以合并的区间
            Interval P = vec[vec.size() - 1];
            if (interval[i].Left <= P.Right + 1) {//两个区间可以合并
                int Left, Right;    
                Right = max(P.Right, interval[i].Right);                
                vec.pop_back();
                Left = P.Left;
                vec.push_back(Interval(Left, Right));
            }
            else {
                vec.push_back(interval[i]);
            }
        }
        int N = vec.size();//区间的数量
    
        for (int i = 1; i <= N;i++) {//c[i],第i个区间为止所有区间的大小的和
            c[i] += c[i-1] + vec[i-1].Right - vec[i-1].Left + 1;
        }

        int s = 1, t = 1;
        ll res = 0;
        ll num;
        for (;;) {
            while (t <= N&&(num=(vec[t-1].Right-vec[s-1].Left+1-(c[t]-c[s-1])) )<= m) {
                ll diff = vec[t - 1].Right - vec[s - 1].Left + 1;
                res =max(res,diff);
                t++;
            }
            if (num <= m) { break; }
            ll remain = c[t-1]-c[s-1]+m ;
            res = max(res, remain);
            s++;
        }        
        res = max(res, c[N ] - c[s - 1] + m);
        printf("%lld\n", res);
    }
    return 0;
}