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>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=m-1; i>=n; i--)
#define SORT(v,n) do{sort(v,v+n);reverse(v,v+n);}while(0)
#define all(x) (x).begin(),(x).end()
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
typedef pair<ll, ll> LP;

ll POW(ll x,ll n){
  x%=MOD;
  if(n==0)return 1;
  if(n%2==0)return POW(x*x,n/2)%MOD;
  return x%MOD*POW(x,n-1)%MOD;
}

ll POW2(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW2(x*x,n/2);
  return x*POW2(x,n-1);
}

ll POW3(ll x,ll n,ll m){
  x%=m;
  if(n==0)return 1;
  if(n%2==0)return POW3(x*x,n/2,m)%m;
  return x*POW3(x,n-1,m)%m;
}

ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}

ll lcm(ll u, ll v) {
  return u*v/gcd(u,v);
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll KAI2(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI2(m-1);
}

ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI(m)%MOD*modinv(KAI(n)%MOD*KAI(m-n)%MOD)%MOD;
}

ll COM2(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI2(m)/KAI2(n)/KAI2(m-n);
}

ll DEC(ll x,ll m,ll n)//xのm進数でのx^nの位の値
{
  return x%POW2(m,n+1)/POW2(m,n);
}

ll keta(ll x,ll n)//xのn進数での桁数
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}

ll DIV(ll x,ll n)//x!のnで割り切れる回数
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}

ll ORD(ll x,ll n)//xのnで割り切れる回数
{
  if(x==0)return INF;
  if(x%n!=0)return 0;
  return 1+ORD(x/n,n);
}

ll SGS(ll x,ll y, ll m)//1+x+…+x^(y-1)をmで割った余り
{
  if(y==0)return 0;
  if(y%2==0){
    return (1+POW3(x,y/2,m))*SGS(x,y/2,m)%m;
  }
  return (1+x*SGS(x,y-1,m))%m;
}

ll SSGS(ll x,ll y,ll m)//Σ[k=1→y](1+x+…+x^(k-1))をmで割った余り
{
  if(y==0)return 0;
  if(y==1)return 1;
  if(y%2==0){
    return (SSGS(x,y/2,m)*(POW3(x,y/2,m)+1)%m+SGS(x,y/2,m)*y/2%m)%m;
  }
  return (SSGS(x,y-1,m)*x%m+y)%m;
}

struct UnionFind
{
  vector<int> par;
  vector<int> sizes;
  UnionFind(int n) : par(n), sizes(n, 1) {
    rep(i,n) par[i] = i;
  }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (sizes[x] < sizes[y]) swap(x, y);
    par[y] = x;
    sizes[x] += sizes[y];
  }
  bool same(int x, int y) {
    return find(x) == find(y);
  }
  int size(int x) {
    return sizes[find(x)];
  }
};

int main()
{
  ll n,k;
  cin >> n >> k;
  FOR(i,k-n+1,n+k){
    printf("%lld ",i);
  }
}