D. Too Easy Problems time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

You are preparing for an exam on scheduling theory. The exam will last for exactly T milliseconds and will consist of n problems. You can either solve problem i in exactly ti milliseconds or ignore it and spend no time. You don't need time to rest after solving a problem, either.

Unfortunately, your teacher considers some of the problems too easy for you. Thus, he assigned an integer a

i to every problem imeaning that the problem i can bring you a point to the final score only in case you have solved no more than ai problems overall (including problem i).

Formally, suppose you solve problems p1, p2, ..., pk during the exam. Then, your final score s will be equal to the number of values of j

between 1 and k such that k ≤ apj.

You have guessed that the real first problem of the exam is already in front of you. Therefore, you want to choose a set of problems to solve during the exam maximizing your final score in advance. Don't forget that the exam is limited in time, and you must have enough time to solve all chosen problems. If there exist different sets of problems leading to the maximum final score, any of them will do.

Input

The first line contains two integers n and T (1 ≤ n ≤ 2·105; 1 ≤ T ≤ 109) — the number of problems in the exam and the length of the exam in milliseconds, respectively.

Each of the next n lines contains two integers ai and ti (1 ≤ ai ≤ n; 1 ≤ ti ≤ 104). The problems are numbered from 1 to n.

Output

In the first line, output a single integer s — your maximum possible final score.

In the second line, output a single integer k (0 ≤ k ≤ n) — the number of problems you should solve.

In the third line, output k distinct integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the indexes of problems you should solve, in any order.

If there are several optimal sets of problems, you may output any of them.

Examples input

5 300
3 100
4 150
4 80
2 90
2 300

output

2
3
3 1 4

input

2 100
1 787
2 788

output

0
0

input

2 100
2 42
2 58

output

2
2
1 2

Note

In the first example, you should solve problems 3, 1, and 4. In this case you'll spend 80 + 100 + 90 = 270 milliseconds, falling within the length of the exam, 300 milliseconds (and even leaving yourself 30 milliseconds to have a rest). Problems 3 and 1 will bring you a point each, while problem 4 won't. You'll score two points.

In the second example, the length of the exam is catastrophically not enough to solve even a single problem.

In the third example, you have just enough time to solve both problems in 42 + 58 = 100 milliseconds and hand your solutions to the teacher with a smile.

Source

My Solution

題意：有m個題目，每個題目有個需要花費的時間ti，以及ai，表示只要最終過題數不超過ai這個題才count。求最大的過題數以及過了哪些題，多種答案則輸出任一答案。

二分+貪心

首先把題目按照ti的為優先順序排序，時間少的在前面。

然後二分答案，mid表示最終的過題數，check的時候，維護剩餘的時間tmp，用cnt表示已選的題目數量，

對於題目從左向右掃，每次如果該題的ai<= mid 則 tmp -= ti，當tmp >= 0是cnt++，然後tmp <= 0時break。

之後cnt >= mid則返回真，否則返回假。

最後得出ans後再按照check的方法再跑一邊就可以得到具體選的題目。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> ii;
const int MAXN = 2e5 + 8;

struct p{
    int t, a, ind;
} ta[MAXN];
inline bool cmp(const p &a, const p &b){
    return a.t < b.t;
}


LL n, time, tmp;
inline bool check(LL mid){
    int i, cnt = 0;
    tmp = time;
    for(i = 0; i < n; i++){
        if(ta[i].a >= mid){
            tmp -= ta[i].t;
            if(tmp >= 0) cnt++;
            else break;
        }
    }
    if(cnt >= mid) return true;
    else return false;

}

vector<int> ansset;
inline bool findans(LL mid){
    int i, cnt = 0;
    tmp = time;
    for(i = 0; i < n; i++){
        if(ta[i].a >= mid){
            tmp -= ta[i].t;
            if(tmp >= 0) cnt++, ansset.push_back(ta[i].ind);
            else break;
        }
        if(cnt == mid) break;
    }
}

int main()
{
    #ifdef LOCAL
    freopen("d.txt", "r", stdin);
    //freopen("d.out", "w", stdout);
    int T = 4;
    while(T--){
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    LL i, a, t, ans = 0;
    cin >> n >> time;
    for(i = 0; i < n; i++){
        cin >> ta[i].a >> ta[i].t;
        ta[i].ind = i+1;

    }
    sort(ta, ta + n, cmp);
    LL l = 0, r = n + 1, mid;
    while(l + 1 < r){
        mid = (l + r) >> 1;
        if(check(mid)){
            ans = mid;
            l = mid;
        }
        else r = mid;
    }

    cout << ans << endl;
    cout << ans << endl;
    findans(ans);
    for(i = 0; i < ans; i++){
        if(i != 0) cout << " ";
        cout << ansset[i];
    }
    cout << endl;


    #ifdef LOCAL
    ansset.clear();
    cout << endl;
    }
    #endif // LOCAL
    return 0;
}

  Thank you!