Given a sequence of N pairs of integers, find the length of the longest increasing subsequence of it.

An increasing sequence A₁..A_n is a sequence such that for every i < j, A_i < A_j.

A subsequence of a sequence is a sequence that appears in the same relative order, but not necessarily contiguous.

A pair of integers (x₁, y₁) is less than (x₂, y₂) iff x₁ < x₂ and y₁ < y₂.

Input

The first line of input contains an integer N (2 ≤ N ≤ 100000).

The following N lines consist of N pairs of integers (x_i, y_i) (-10⁹ ≤ x_i, y_i ≤ 10⁹).

Output

The output contains an integer: the length of the longest increasing subsequence of the given sequence.

Example

Input:
8
1 3
3 2
1 1
4 5
6 3
9 9
8 7
7 6

Output:
3

題意；求最長的序列LIS，滿足i<j，xi<xj ；yi<yj。

思路：裸題，cqd分治，計算每個[L,Mid]區間對[Mid+1,R]區間的貢獻。

有兩個註意點：

第一：由於時間緊，只有300ms，不能寫結構體的排序； 這裏借鑒了別人的一維排序（ORZ，強的啊）。

第二：註意規避x1=x2，y1<y2的時候不能用 1去更新2。（和求逆序對那題一樣，只有把y坐標的放左邊即可）。

#include<bits/stdc++.h>
using
 
 namespace std;
const int maxn=1000010;
int p[maxn],a[maxn],b[maxn],dp[maxn],Mx[maxn],tot,ans;
bool cmp(int x,int y){ if(a[x]==a[y]) return x>y; return a[x]<a[y]; }
void solve(int L,int R)
{
    if(L==R){ dp[L]=max(dp[L],1); return ;}
    int Mid=(L+R)/2;
    solve(L,Mid); 
    for(int i=L;i<=R;i++) p[i]=i;
    sort(p+L,p+R+1,cmp);
    for(int i=L;i<=R;i++){
        if(p[i]<=Mid) for(int j=b[p[i]];j<=tot;j+=(-j)&j) Mx[j]=max(Mx[j],dp[p[i]]);
        else for(int j=b[p[i]]-1;j;j-=(-j)&j) dp[p[i]]=max(dp[p[i]],Mx[j]+1); 
    }
    for(int i=L;i<=R;i++)
      if(p[i]<=Mid) for(int j=b[p[i]];j<=tot;j+=(-j)&j) Mx[j]=0;
    solve(Mid+1,R);
}
int main()
{
    int N,i,fcy=0;
    scanf("%d",&N);
    for(i=1;i<=N;i++) scanf("%d%d",&a[i],&b[i]),p[i]=b[i];
    sort(p+1,p+N+1);
    tot=unique(p+1,p+N+1)-(p+1);
    for(i=1;i<=N;i++) b[i]=lower_bound(p+1,p+tot+1,b[i])-p;
    solve(1,N);
    for(i=1;i<=N;i++) fcy=max(fcy,dp[i]);
    printf("%d\n",fcy);
    return 0;
}

SPOJ:Another Longest Increasing Subsequence Problem(CDQ分治求三維偏序)