Codeforces Round #240 (Div. 2) Mashmokh and Numbers

题意找出一组数列，满足每相邻的两个数的最大公因数和为给定的值。，注意相邻两个数的公约数为一就好做了，注意下边界即可

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, ..., an 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, ..., an (1 ≤ ai ≤ 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.

#include <stdio.h>int main(){    int n,k;    while(scanf("%d %d",&n,&k)==2)    {        if(n==1&&k!=0) printf("-1\n");        else if(n==1&&k==0) printf("1\n");        else if(n/2>k) printf("-1\n");        else        {            int a=1,b=2;            int t=n/2; t*=2; int x=n/2;            int last=k-(x-1);            for(int i=1;i<x;i++)            {               if(i==1)               {                   while(a==last||b==last||a==2*last||b==2*last)a+=2,b+=2;                   printf("%d %d",a,b);               }               else               {                    while(a==last||b==last||a==2*last||b==2*last)a+=2,b+=2;                    printf(" %d %d",a,b);               }               a+=2,b+=2;            }            if(n==2)printf("%d %d",last,2*last);            else printf(" %d %d",last,2*last);            if(n&1)printf(" 1000000000");            printf("\n");        }    }    return 0;}