Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the rings for this wonderful game? Once upon a time, the ruler of the ancient Egypt ordered the workers of Hanoi Factory to create as high tower as possible. They were not ready to serve such a strange order so they had to create this new tower using already produced rings.

There are n rings in factory's stock. The i-th ring has inner radius ai, outer radius bi and height hi. The goal is to select some subset of rings and arrange them such that the following conditions are satisfied:

• Outer radiuses form a non-increasing sequence, i.e. one can put the j-th ring on the i-th ring only if bj ≤ bi.
• Rings should not fall one into the the other. That means one can place ring j on the ring i only if bj > ai.
• The total height of all rings used should be maximum possible.

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of rings in factory's stock.

The i-th of the next n lines contains three integers ai, bi and hi (1 ≤ ai, bi, hi ≤ 109, bi > ai) — inner radius, outer radius and the height of the i-th ring respectively.

Print one integer — the maximum height of the tower that can be obtained.

In the first sample, the optimal solution is to take all the rings and put them on each other in order 3, 2, 1.

In the second sample, one can put the ring 3 on the ring 4 and get the tower of height 3, or put the ring 1 on the ring 2 and get the tower of height 4.

因a和b数组中的数较大，因此需要离散化。可以用类似于LIS的方法进行dp转移，但因为题目要求时间复杂度为O（nlogn），所以还要用树状数组或线段树进行优化，维护1到某个半径的最大高度。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<algorithm>

#include<cstring>

#include<string>

#include<vector>

#include<map>

#include<set>

#include<queue>

using namespace std;

struct ss

{

    long long a,b,c;

};

ss a[];

long long n,len,tree[],m=;

struct Hash : vector<int> {    //离散化

    void prepare() {

        sort(begin(), end());

        //erase(unique(begin(), end()), end());

    }

    long long get(long long x) {

        return lower_bound(begin(), end(), x)-begin()+;

    }

} has;

void upd(long long x,long long y)

{

    for (;x<=m;x+=x&(-x)) tree[x]=max(tree[x],y);

}

long long sum(long long x)

{

    long long p=;

    for (;x;x-=x&(-x)) p=max(p,tree[x]);

    return p;

}

inline bool cmp(ss a,ss b)

{

    return (a.b>b.b||a.b==b.b&&a.a>b.a);

}

int main()

{

    scanf("%lld",&n);

    long long i;

    for (i=;i<=n;i++)

        scanf("%lld%lld%lld",&a[i].a,&a[i].b,&a[i].c);

    has.clear();

    for (i=;i<=n;i++)

        has.push_back(a[i].a),has.push_back(a[i].b);

    has.prepare();

    for (i=;i<=n;i++)

        a[i].a=has.get(a[i].a),a[i].b=has.get(a[i].b),m=max(a[i].b,m);

    m*=;

    sort(a+,a+n+,cmp);

    //for (i=1;i<=n;i++)

    //    printf("%d %d %d\n",a[i].a,a[i].b,a[i].c);

    //cout<<m<<endl;

    memset(tree,,sizeof(tree));

    long long ans=;

    for (i=;i<=n;i++)

    {

        long long now=sum(a[i].b-)+a[i].c;

        //cout<<sum(a[i].b-1)<<endl;

        upd(a[i].a,now);

        ans=max(ans,now);

    }

    printf("%lld\n",ans);

    return ;

}