Coffee and Coursework (Hard Version)

Coffee and Coursework (Hard Version)

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The only difference between easy and hard versions is the constraints.

Polycarp has to write a coursework. The coursework consists of mm pages.

Polycarp also has nn cups of coffee. The coffee in the ii -th cup Polycarp has aiai caffeine in it. Polycarp can drink some cups of coffee (each one no more than once). He can drink cups in any order. Polycarp drinks each cup instantly and completely (i.e. he cannot split any cup into several days).

Surely, courseworks are not being written in a single day (in a perfect world of Berland, at least).

Let's consider some day of Polycarp's work. Consider Polycarp drinks kk cups of coffee during this day and caffeine dosages of cups Polycarp drink during this day are ai1,ai2,…,aikai1,ai2,…,aik . Then the first cup he drinks gives him energy to write ai1ai1 pages of coursework, the second cup gives him energy to write max(0,ai2−1)max(0,ai2−1) pages, the third cup gives him energy to write max(0,ai3−2)max(0,ai3−2) pages, ..., the kk -th cup gives him energy to write max(0,aik−k+1)max(0,aik−k+1) pages.

If Polycarp doesn't drink coffee during some day, he cannot write coursework at all that day.

Polycarp has to finish his coursework as soon as possible (spend the minimum number of days to do it). Your task is to find out this number of days or say that it is impossible.

Input

The first line of the input contains two integers nn and mm (1≤n≤2⋅1051≤n≤2⋅105 , 1≤m≤1091≤m≤109 ) — the number of cups of coffee and the number of pages in the coursework.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109 ), where aiai is the caffeine dosage of coffee in the ii -th cup.

Output

If it is impossible to write the coursework, print -1. Otherwise print the minimum number of days Polycarp needs to do it.

Examples

Input

Copy

5 8
2 3 1 1 2

Output

Copy

4

Input

Copy

7 10
1 3 4 2 1 4 2

Output

Copy

2

Input

Copy

5 15
5 5 5 5 5

Output

Copy

1

Input

Copy

5 16
5 5 5 5 5

Output

Copy

2

Input

Copy

5 26
5 5 5 5 5

Output

Copy

-1

Note

In the first example Polycarp can drink fourth cup during first day (and write 11 page), first and second cups during second day (and write 2+(3−1)=42+(3−1)=4 pages), fifth cup during the third day (and write 22 pages) and third cup during the fourth day (and write 11 page) so the answer is 44 . It is obvious that there is no way to write the coursework in three or less days.

In the second example Polycarp can drink third, fourth and second cups during first day (and write 4+(2−1)+(3−2)=64+(2−1)+(3−2)=6 pages) and sixth cup during second day (and write 44 pages) so the answer is 22 . It is obvious that Polycarp cannot write the whole coursework in one day in this test.

In the third example Polycarp can drink all cups of coffee during first day and write 5+(5−1)+(5−2)+(5−3)+(5−4)=155+(5−1)+(5−2)+(5−3)+(5−4)=15 pages of coursework.

In the fourth example Polycarp cannot drink all cups during first day and should drink one of them during the second day. So during first day he will write 5+(5−1)+(5−2)+(5−3)=145+(5−1)+(5−2)+(5−3)=14 pages of coursework and during second day he will write 55 pages of coursework. This is enough to complete it.

In the fifth example Polycarp cannot write the whole coursework at all, even if he will drink one cup of coffee during each day, so the answer is -1.

分析：大数据，既然已经使用sort排序了，直接二分就可以了鸭

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 #define ll long long
 int n,m;
 vector<int> a;

 bool can(int i){
     ll sum=;
     for( int j=; j<n; j++ ){
         sum+=max(a[j]-j/i,);
     }
     if(sum>=m)return true;
     else return false;
 }

 int main(int argc, char const *argv[])
 {
     cin>>n>>m;
     a=vector<int>(n);

     for( int i=; i<n; i++ ){
         cin>>a[i];
     }

     sort(a.rbegin(),a.rend());

     int l=,r=n;
     while(l<=r){
         ll mid=(l+r) >> ;
         if(can(mid)) r=mid-;
         else l=mid+;
     }

     if(can(l)) cout<<l<<endl;
     // else if(can(r)) cout<<r<<endl;
     else{
         cout<<-<<endl;
     }

     return ;
 }