CF---C. Mashmokh and Numbers

C. Mashmokh and Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.

In this game Mashmokh writes sequence of n distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers x and y from the board, he gets gcd(x, y) points. At the beginning of the game Bimokh has zero points.

Mashmokh wants to win in the game. For this reason he wants his boss to get exactly k points in total. But the guy doesn't know how choose the initial sequence in the right way.

Please, help him. Find n distinct integers a₁, a₂, ..., a_n such that his boss will score exactly k points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 10⁹.

Input

The first line of input contains two space-separated integers n, k (1 ≤ n ≤ 10⁵; 0 ≤ k ≤ 10⁸).

Output

If such sequence doesn't exist output -1 otherwise output n distinct space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Sample test(s)

input

5 2

output

1 2 3 4 5

input

5 3

output

2 4 3 7 1

input

7 2

output

-1

Note

gcd(x, y) is greatest common divisor of x and y.

 

 

//这道题目WA了很多次，一开始思路是对的，还是需要从题目中寻找好的突破口， 这里k的取值范围比较特殊，不到10^9，而是10^8.

突破口就在这里！ 但是后来注释那里分类有点错误，WA了很多次，这里做一总结，还是不够细心。

Code：

#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>#include <iomanip>#include <iostream>using namespace std; const int N=100000001; int main(){    freopen("in.txt","r",stdin);    int i,j,k,n,m, a,b;    while(scanf("%d %d",&n,&m)!=EOF){        if(n==1 && m!=0) printf("-1\n");        else{            if(n/2 > m) printf("-1\n");            else if(n/2 == m){                printf("1");                for(i=2;i<=n;i++) printf(" %d",i);                printf("\n");            }            else{                int t=n/2;                k=m-t+1;                                int kk=N%k;                kk=k-kk;                printf("%d %d",k,N+kk);                                  m=n-2;                k=100001;   //这里一开始枚举的位置出错了，导致出现了重复！                 while(m--){                    printf(" %d",k);                    k++;                }                printf("\n");            }        }    }        return 0;}