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++ (GCC 9.2.1) (judge verdict: Accepted)

/*
─────────────────────
───────────████████──
──────────███▄███████
──────────███████████
──────────███████████
──────────██████─────
──────────█████████──
█───────███████──────
██────████████████───
███──██████████──█───
███████████████──────
███████████████──────
─█████████████───────
──███████████────────
────████████─────────
─────███──██─────────
─────██────█─────────
─────█─────█─────────
─────██────██────────
─────────────────────
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>

#define int long long
#define read(a) int a; cin >> a;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define reads(s) string s; cin >> s;
#define readarr(a, n) int a[(n) + 1] = {}; fori (i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; fori (i, 1, n) {fori (j, 1, m) cin >> a[i][j];}

#define print(a) cout << a << endl;
#define printarr(a, n) fori (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) fori (i, 1, n) {fori (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(s) (int)(s.size())
#define pb push_back
#define fi first
#define se second
#define rz resize
#define fori(i, a, b) for (int i = (a); i <= (b); i++)
#define ford(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
#define setprec(x) cout << fixed << setprecision(x);
 
using namespace std;
using namespace __gnu_pbds; 
 
typedef tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pair <int, int>> vpi;
typedef vector < vector <int> > vvi;
 
const int INF = 4e18;
const int mod = 1e9+7; 

     
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);


    //start from here

    readb(k, x)

    int l = -1e6, r = 1e6;

    fori (i, max(x-k+1, l), min(r, x+k-1)) cout << i << " ";

    return 0;
}