ASC(1)G（上升时间最长的序列）

G - Beautiful People

Time Limit: 10000/5000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)    
Special Judge

SubmitStatus

Problem Description

The most prestigious sports club in one city has exactly N members. Each of its members is strong and beautiful. More precisely, i-th member of this club (members being numbered by the time they entered the
club) has strength Si and beauty Bi. Since this is a very prestigious club, its members are very rich and therefore extraordinary people, so they often extremely hate each other. Strictly speaking, i-th member of the club Mr X hates j-th member of the club
Mr Y if Si <= Sj and Bi >= Bj or if Si >= Sj and Bi <= Bj (if both properties of Mr X are greater then corresponding properties of Mr Y, he doesn't even notice him, on the other hand, if both of his properties are less, he respects Mr Y very much).

To celebrate a new 2003 year, the administration of the club is planning to organize a party. However they are afraid that if two people who hate each other would simultaneouly attend the party, after a drink
or two they would start a fight. So no two people who hate each other should be invited. On the other hand, to keep the club prestige at the apropriate level, administration wants to invite as many people as possible.

Being the only one among administration who is not afraid of touching a computer, you are to write a program which would find out whom to invite to the party.

Input

      The first line of the input file contains integer N — the number of members of the club. (2 ≤ N ≤ 100 000). Next N lines contain two numbers each — Si and
Bi respectively (1 ≤ Si, Bi ≤
109).

Output

      On the first line of the output file print the maximum number of the people that can be invited to the party. On the second line output N integers — numbers of members to be invited in arbitrary order. If several solutions exist, output any one.

Sample Input

4
1 1
1 2
2 1
2 2

Sample Output

2
1 4

题意：给了n个数对x,y。要求找出最长的数对。使得每一维都严格递增

思路：最长上升子序列，仅仅只是是2维的。能够先按x排序以后按顺序用线段树维护就好了

            详细维护的是，以每一个y点结尾的最大长度

            需要注意的是，当相同的第一更新x在下面y在记录后，所有查询，更新统一。由于x点是不等于所述增加的序列中可以是足够