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 <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using namespace boost::multiprecision;

# define REP(i,n) for (int i=0;i<(n);++i)
# define rep(i,a,b) for(int i=a;i<(b);++i)
# define p(s) std::cout << s ;
# define pl(s)  std::cout << s << endl;
# define all(v) v.begin(),v.end()
# define showVector(v) REP(i,v.size()){p(v[i]);p(" ")} pl("")
template<class T> inline bool chmin(T &a, T b){ if(a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){ if(a < b) { a = b; return true;} return false;}
typedef long long int ll;
typedef pair<ll,ll> P_ii;
typedef pair<double,double> P_dd;

template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}

template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}

ll gcd(ll a, ll b) {
    if(a < b) swap(a,b);
    
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b){
    ll g = gcd(a,b);
    return (a/g)*b;
}

// 素数判定 O(√n)
bool is_prime(int n){
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0) return false;
    }
    return true;
}

// 約数列挙 O(√n)
vector<ll> divisor(ll n){
    vector<ll> res;
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n / i) res.push_back(n / i);
        }
    }
    return res;
}

vector<pair<ll, ll> > prime_factorize(ll n) {
    vector<pair<ll, ll> > res;
    for (ll p = 2; p * p <= n; ++p) {
        if (n % p != 0) continue;
        ll num = 0;
        while (n % p == 0) { ++num; n /= p; }
        res.push_back(make_pair(p, num));
    }
    if (n != 1) res.push_back(make_pair(n, 1));
    return res;
}

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x(x%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

const int MOD = 1000000007;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;

void addM(ll &a, ll b) {
    a += b;
    if (a >= MOD) a -= MOD;
}

void mulM(ll &a, ll b) {
    a = ((a%MOD)*(b%MOD))%MOD ;
}

ll powM(ll a,ll b) {
    ll ret = 1;
    ll tmp = a;
    while(b>0) {
        if((b&1)==1) ret = (ret * tmp) % MOD;
        tmp = (tmp * tmp) % MOD;
        b = b >> 1;
    }
    return ret;
}

// mod. m での a の逆元 a^{-1} を計算する
ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);  

  int K, X;
  cin >> K >> X;

  for(int i = X - (K - 1); i <= X + (K - 1); i++){
    cout << (i == X - (K - 1) ? "" : " ") << i;
  }
  cout << endl;


  return 0;
}