https://codeforces.com/contest/1305/problem/E
The meaning of problems: the number of configuration requirements n satisfy m equilibrium, balancing rule: (I, J, K). 1 <= I <J <K <&& n = A I + A J = A K .
Solution: structure 1, 2, 3 ..... n apparent that the maximum number of sequences have to balance the number of each may be understood Contribution (i - 1) / 2 balanced, it can be considered as the maximum number of equilibrium.
Balance can not be greater than the maximum number of the construct sequence.
Traversal: If the number m of current required to balance> = (i-1) / 2, then ai = i.
Otherwise, the number (i-1) * 2 + 1 - 2 * (m), this time to meet the balance has reached the required number, the number of contributions no longer balanced.
If after that there was a number 1000000000- (ni) * 10000, the number of contributions to avoid balance.
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define cin() scanf("%lld" , &x);
using namespace std;
const int esp = 1e-6;
const int maxn = 5e5+5;
void solve(){
int n , m ;
cin >> n >> m ;
int ma = ((1 + (n-1)/2)*((n-1)/2))/2;
if(n % 2 == 1){
ma *= 2 ;
ma -= (n-1)/2;
}else ma *= 2 ;
if(m > ma){
cout << -1 << endl;
return ;
}else{
rep(i , 1 , n){
if(m == 0){
cout << 100000000 - (n-i)*10000 << " " ;
}else if(m >= (i-1)/2){
cout << i << " " ;
m -= (i-1)/2 ;
}else{
cout << 2*(i-1)+1 - 2*(m) << " " ;
m = 0 ;
}
}
}
}
signed main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}