In a galaxy far, far away Lesha the student has just got to know that he has an exam in two days. As always, he hasn't attended any single class during the previous year, so he decided to spend the remaining time wisely.

Lesha knows that today he can study for at most aa hours, and he will have bb hours to study tomorrow. Note that it is possible that on his planet there are more hours in a day than on Earth. Lesha knows that the quality of his knowledge will only depend on the number of lecture notes he will read. He has access to an infinite number of notes that are enumerated with positive integers, but he knows that he can read the first note in one hour, the second note in two hours and so on. In other words, Lesha can read the note with number kk in kkhours. Lesha can read the notes in arbitrary order, however, he can't start reading a note in the first day and finish its reading in the second day.

Thus, the student has to fully read several lecture notes today, spending at most aa hours in total, and fully read several lecture notes tomorrow, spending at most bb hours in total. What is the maximum number of notes Lesha can read in the remaining time? Which notes should he read in the first day, and which — in the second?

The only line of input contains two integers aa and bb (0≤a,b≤1090≤a,b≤109) — the number of hours Lesha has today and the number of hours Lesha has tomorrow.

In the first line print a single integer nn (0≤n≤a0≤n≤a) — the number of lecture notes Lesha has to read in the first day. In the second line print nn distinct integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤a1≤pi≤a), the sum of all pipi should not exceed aa.

In the third line print a single integer mm (0≤m≤b0≤m≤b) — the number of lecture notes Lesha has to read in the second day. In the fourth line print mm distinct integers q1,q2,…,qmq1,q2,…,qm (1≤qi≤b1≤qi≤b), the sum of all qiqi should not exceed bb.

All integers pipi and qiqi should be distinct. The sum n+mn+m should be largest possible.

In the first example Lesha can read the third note in 33 hours in the first day, and the first and the second notes in one and two hours correspondingly in the second day, spending 33 hours as well. Note that Lesha can make it the other way round, reading the first and the second notes in the first day and the third note in the second day.

In the second example Lesha should read the third and the sixth notes in the first day, spending 99 hours in total. In the second day Lesha should read the first, second fourth and fifth notes, spending 1212 hours in total.

题意：

　有无数多个lecture note，标号为1，2，3，4，5... 他们复习所要花费的时间和标号大小相同。分两天复习这些lecture。每天的复习小时不一样，求分配使得复习的lecture最多。

思路：

首先a,b分开装东西，不会比他们合在一起装东西装的多。（反证法）

其次，如果从大到小枚举可行的lecture note，如果每次先考虑a，那么a最后肯定可以被装满。

所以，剩下的lecture note，给b，b一定装的下所有。

代码：

 #include <iostream>

 #include <cstring>

 #include <algorithm>

 #include <cmath>

 #include <cstdio>

 #include <vector>

 #include <queue>

 #include <set>

 #include <map>

 #include <stack>

 #define ll long long

 #define local

 using namespace std;

 const int MOD = 1e9+;

 const int inf = 0x3f3f3f3f;

 const double PI = acos(-1.0);

 const int maxn = (1e5+);

 const int maxedge = *;

 ll a, b;

 int main() {

 #ifdef local

     if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");

 #endif

     scanf("%lld%lld", &a, &b);

     vector <ll> ansa, ansb;

     ll x = ;

     for (; 1LL*x*(x+)/ <= a+b; ++x) ;

     x--;

     for (; x > ; x--) {

         if (a >= x) {

             ansa.push_back(x);

             a -= x;

         } else if (b >= x) {

             ansb.push_back(x);

             b -= x;

         }

     }

     printf("%d\n", int(ansa.size()));

     for (int i = ; i < ansa.size(); ++i) {

         printf("%lld", ansa[i]);

         if (i != ansa.size()-) printf(" ");

     }

     cout << endl;

     printf("%d\n", int(ansb.size()));

     for (int i = ; i < ansb.size(); ++i) {

         printf("%lld", ansb[i]);

         if (i != ansb.size()-) printf(" ");

     }

     cout << endl;

 #ifdef local

     fclose(stdin);

 #endif

     return ;

 }