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: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <string>
#include <queue>
#include <cmath>
#include <iomanip>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <random>
#include <unordered_map>
#include <unordered_set>
#define ll long long
#define fr(i,k,N) for(ll i=k; i<N;i++)
#define rep(i,N) for(ll i=0; i<N;i++)
#define ALL(v) v.begin(),v.end()
#define mp make_pair
using namespace std;
template <class X> void pr(X test){cout<<test<<endl;}
template <class X> void prr(X test){for(auto it:test){cout<<it<<endl;}}
template <class X> void prrr(X test){int f=0;for(auto it:test){cout<<(f++!=0?" ":"")<<it;}cout<<endl;}
template<typename X>
istream&operator>>(istream&i,vector<X>&v){for(X&x:v)i>>x;return i;}

class UF { // union find from https://github.com/kartikkukreja/blog-codes (MIT)
    // [usage]
    // auto uf = new UF(n);
    // uf->merge(l, r); // 0 <= l, r <= n-1
    // ...
public:
    int *id; // inner id
    int cnt; // number of all vartices
    int *sz; // sz[i] means size of i's set
    UF(int N)   {
        cnt = N;
        id = new int[N];
        sz = new int[N];
        for(int i=0; i<N; i++) {
            id[i] = i;
            sz[i] = 1;
        }
    }
    ~UF() {
        delete [] id;
        delete [] sz;
    }
    int find(int p) {
        int root = p;
        while (root != id[root]) root = id[root];
        while (p != root) {
            int newp = id[p];
            id[p] = root;
            p = newp;
        }
        return root;
    }
    void merge(int x, int y) {
        int i = find(x);
        int j = find(y);
        if (i == j) return;
        
        if (sz[i] < sz[j]) {
            id[i] = j;
            sz[j] += sz[i];
        } else {
            id[j] = i;
            sz[i] += sz[j];
        }
        cnt--;
    }
    bool connected(int x, int y) {
        return find(x) == find(y);
    }
    int count() {
        return cnt;
    }
};

// return sorted 約数list 1 <= x <= n
vector<ll> divv(ll n) {
  vector<ll> v;
  for(ll i = 1; i*i <= n; i++) {
    if(n%i == 0) {
      v.push_back(i);
      if(i*i != n) v.push_back(n/i);
    }
  }
  sort(ALL(v));
  return v;
}

int main() {
    int h,w;
    string s;
    cin>>h>>w;
    auto f = false;
rep(i, 2000001){
    auto y = i-1000000;
    if(abs(y - w) < h) {
        if(f) cout<<" ";
        f=true;
        cout<<y;
    }

}

    pr("");
    return 0;
}