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//
#include <bits/stdc++.h>
#define int long long
#define ll long long 
#define rep(i,a,b) for(signed i=a;i<(b);++i)
#define erep(i,a,b) for(signed i=a;i<=(b);++i)
#define per(i,a,b) for(signed i=(a);i>(b);--i)
#define eper(i,a,b) for(signed i=(a);i>=b;--i)
#define fore(i, x, a) for(auto &&x:a) 
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define pb push_back
#define mp make_pair
#define ALL(x) begin(x),end(x)
#define F first
#define S second
#define debug(x)  cerr << #x << ": " << (x) << '\n';
const long long INF=1001001001001001001;
const int MOD=(int)1e9 + 7;
const double EPS=1e-9;
using namespace std;
using Pii = pair<int,int>;
using vii = vector<int>;
template<class T>using  PS_queue = priority_queue<T, vector<T>, greater<T> >;
template<class T>using vv = vector<T>;
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<typename A, size_t N, typename T> void Fill(A (&array)[N], const T &val){fill( (T*)array, (T*)(array+N), val );}
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v))os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T>T gcd(T a, T b){ return b ? gcd(b, a % b) : a; }
template<class T>T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct edge { int from, to, cost; };
int dy[]={0, 1, -1, 0}; int dx[]={1, 0, 0, -1};
// cout << fixed;
// cout << setprecision(10) << val;

int n, m;
signed main() {
 cin.tie(0);
 ios::sync_with_stdio(false);
  cin >> n >> m;
  vii a(2*n+1), b(2*n+1);
  iota(ALL(a), 0);
  erep(i, 1, m) {
    int k; cin >> k;
    if (k == 0) {
      erep(j, 1, n) {
        b[2*j-1] = a[j];
      } 
      erep(j, n+1, 2*n) {
        b[2*(j-n)] = a[j];
      }
    } else {
      erep(j, 1, k) {
        b[2*n-k+j] = a[j];
      } 
      erep(j, k+1, 2*n) {
        b[j-k] = a[j];
      }
    }
    erep(j, 1, 2*n) {
      a[j] = b[j];
    }
  }
  erep(i, 1, 2*n) cout << a[i] << endl;
  return 0;
}




Measured memory_bytes: 3188