Problem:
Score : 200 points

Problem Statement

There are 2000001 stones placed on a number line. The coordinates of these stones are -1000000, -999999, -999998, \ldots, 999999, 1000000.

Among them, some K consecutive stones are painted black, and the others are painted white.

Additionally, we know that the stone at coordinate X is painted black.

Print all coordinates that potentially contain a stone painted black, in ascending order.

Constraints

1 \leq K \leq 100

0 \leq X \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K X

Output

Print all coordinates that potentially contain a stone painted black, in ascending order, with spaces in between.

Sample Input 1

3 7

Sample Output 1

5 6 7 8 9

We know that there are three stones painted black, and the stone at coordinate 7 is painted black. There are three possible cases:

The three stones painted black are placed at coordinates 5, 6, and 7.

The three stones painted black are placed at coordinates 6, 7, and 8.

The three stones painted black are placed at coordinates 7, 8, and 9.

Thus, five coordinates potentially contain a stone painted black: 5, 6, 7, 8, and 9.

Sample Input 2

4 0

Sample Output 2

-3 -2 -1 0 1 2 3

Negative coordinates can also contain a stone painted black.

Sample Input 3

1 100

Sample Output 3

100

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// template
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<long long, long long> LP;
#define INF 999999999
#define MOD 1000000007
#define REP(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define REP_AB(i, a, b) for (ll i = ll(a); i < ll(b); i++)
#define ALL(v) v.begin(), v.end()
#define SORT(v) sort(ALL(v))
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)), v.end());
#define SIZE(x) ((ll)(x).size())
#define REVERSE(v) reverse(ALL(v))
// true / false
#define BIN_SEARCH(v, a) binary_search(ALL(v), a)
// index
#define BIN_LEFT(v, a) (lower_bound(ALL(v), a) - v.begin())
#define BIN_RIGHT(v, a) (upper_bound(ALL(v), a) - v.begin())
#define BIN_INSERT(v, a) (v.insert(v.begin() + BIN_LEFT(v, a), a))
#define DEL(v, i) v.erase(v.begin() + i)
#define INSERT(v, i, a) v.insert(v.begin() + i, a)
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return (a / gcd(a, b)) * b;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
#define out cout
#define in cin
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
// template end
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    ll K, X;
    in >> K >> X;
    //vector<ll> X(N)

    REP_AB(i, X-K+1, X+K){
        if(i==1000001){
            break;
        }else if(i<-1000001){
            i=-1000000;
        }
        out << i;
    }

}