Problem:
Score : 600 points

Problem Statement

10^9 contestants, numbered 1 to 10^9, will compete in a competition.
There will be two contests in this competition.

The organizer prepared N problems, numbered 1 to N, to use in these contests.
When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.

The organizer will use these N problems in the two contests.
Each problem must be used in exactly one of the contests, and each contest must have at least one problem.

The joyfulness of each contest is the number of contestants who will solve all the problems in the contest.
Find the maximum possible total joyfulness of the two contests.

Constraints

2 \leq N \leq 10^5

1 \leq L_i \leq R_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 R_1
L_2 R_2
\vdots
L_N R_N

Output

Print the maximum possible total joyfulness of the two contests.

Sample Input 1

4
4 7
1 4
5 8
2 5

Sample Output 1

6

The optimal choice is:

Use Problem 1 and 3 in the first contest. Contestant 5, 6, and 7 will solve both of them, so the joyfulness of this contest is 3.

Use Problem 2 and 4 in the second contest. Contestant 2, 3, and 4 will solve both of them, so the joyfulness of this contest is 3.

The total joyfulness of these two contests is 6. We cannot make the total joyfulness greater than 6.

Sample Input 2

4
1 20
2 19
3 18
4 17

Sample Output 2

34

Sample Input 3

10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526

Sample Output 3

540049931

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
* じょえチャンネル
* 高評価・チャンネル登録よろしくおねがいします！
* https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
*/

#include <bits/stdc++.h>

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)

//here!!!
// define int long long !!!!!

#define int long long

// define int long long !!!!!


#define mod (int)((1e9)+7)

#define intt long long
#define itn long long
#define P pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)

#define ALL(v) v.begin(),v.end()

using namespace std;

//Library
//モッドパウ
inline int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % m;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}
//Library-End

class modInt {
    int value, modulo;
public:
    modInt() : value(0), modulo(mod) { value = 0; }
    template<typename T>
    modInt(T value = 0, int modulo = mod) : value(value), modulo(modulo) {
        if (value < 0) {
            value = -value;
            value %= modulo;
            value = -value;
            value += modulo;
        }
        this->value = value % modulo;
    }
    inline operator int()const { return value; }
    inline modInt& operator+=(modInt x) {
        value += x.value;
        if (value >= modulo)value -= modulo;
        return *this;
    }
    inline modInt& operator++() {
        if (value == modulo - 1)value = 0;
        else value++;
        return *this;
    }
    inline modInt& operator-()const {
        return modInt(0) -= *this;
    }
    inline modInt& operator-=(modInt x) {
        value -= x.value;
        if (value < 0)value += modulo;
        return *this;
    }
    inline modInt& operator--() {
        if (value == 0)value = modulo - 1;
        else value--;
        return *this;
    }
    inline modInt& operator*=(modInt x) {
        value = value * x.value % modulo;
        return *this;
    }
    inline modInt& operator/=(modInt x) {
        return operator*=(x.inv());
    }
    inline modInt inv() {
        return modpow(*this, modulo - 2);
    }
    template<typename T> modInt operator+(T x) { return modInt(*this) += x; }
    template<typename T> modInt& operator+=(T x) { return operator+=(modInt(x, modulo)); }
    template<typename T> modInt operator-(T x) { return modInt(*this) -= x; }
    template<typename T> modInt& operator-=(T x) { return operator-=(modInt(x, modulo)); }
    template<typename T> modInt operator*(T x) { return modInt(*this) *= x; }
    template<typename T> modInt& operator*=(T x) { return operator*=(modInt(x, modulo)); }
    template<typename T> modInt operator/(T x) { return modInt(*this) /= x; }
    template<typename T> modInt& operator/=(T x) { return operator/=(modInt(x, modulo)); }
};
istream& operator>>(istream& ist, modInt& x) {
    int a;
    ist >> a;
    x = a;
    return ist;
}
#define vecin(v) for(int i=0;i<v.size();i++)scanf("%lld",&v[i]);
#define vecout(v) {for(int i=0;i<v.size();i++)printf("%lld\n",v[i]);}
#define endl "\n" //interactive の時に注意！！！
//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理
    
    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }
    
public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }
    
    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }
    
    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }
    
    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
    vector<int> ookisa;
    
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        ookisa=vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
            ookisa[i]=1;
        }
    }
    
    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }
    
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
            ookisa[y]+=ookisa[x];
            ookisa[x]=0;
        }
        else {
            par[y] = x;
            ookisa[x]+=ookisa[y];
            ookisa[y]=0;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }
    int size(int i){
        i=find(i);
        return ookisa[i];
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    
};

//Union-Find-End



int n,l[100004],r[100004],lrui1[100004],rrui1[1000004],lrui2[100004],rrui2[100004],ans;
P ls,rs,lb,rb;
signed main(){
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin>>n;
    rs.second=inf;
    lb.first=inf;
    REP(i,n){
        cin>>l[i]>>r[i];
        if (rs.second>r[i]) {
            rs={l[i],r[i]};
        }
        else if (ls.first<l[i]) {
            ls={l[i],r[i]};
        }
        else if (lb.first>l[i]) {
            lb={l[i],r[i]};
        }
        else if (rb.second<r[i]) {
            rb={l[i],r[i]};
        }
    }
    
    rrui1[0]=inf;
    rrui2[n+1]=inf;
    REP(i,n){
        lrui1[i]=max(lrui1[i-1],l[i]);
        rrui1[i]=min(rrui1[i-1],r[i]);
    }
    for (int i=n; i>=1; i--) {
        lrui2[i]=max(lrui2[i+1],l[i]);
        rrui2[i]=min(rrui2[i+1],r[i]);
    }

    
    REP(i,n){
        chmax(ans, max(0ll,min(rrui1[i-1],rrui2[i+1])-max(lrui1[i-1],lrui2[i+1])+1)+(r[i]-l[i]+1));
    }
//    chmax(ans, max(0ll,rb.second-ls.first)+max(0ll,rs.second-lb.first));
    cout<<ans<<endl;
}