#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <chrono>
#define _USE_MATH_DEFINES
#include <cmath>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;

#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()

const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
// const int dy[] = {1, 1, 0, -1, -1, -1, 0, 1},
//           dx[] = {0, -1, -1, -1, 0, 1, 1, 1};

struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    cerr << fixed << setprecision(10);
  }
} iosetup;
/*-------------------------------------------------*/
void output(const vector<bool> &ans) {
  for (bool e : ans) cout << e;
  cout << '\n';
  exit(0);
}

int main() {
  int n, x; cin >> n >> x;
  vector<int> a(n); REP(i, n) cin >> a[i];
  vector<bool> ans(x, false);
  if (n == 1) {
    if (a[0] <= x) ans[a[0] - 1] = true;
    output(ans);
  }
  if (n >= 19) output(ans);
  long long lb = 1;
  REP(i, n) {
    lb *= a[i] + 1;
    if (x < lb - 1) output(ans);
  }
  if (count(ALL(a), 1) > 0) {
    FOR(i, lb - 1 - 1, x) ans[i] = true;
    output(ans);
  }
  vector<int> p(n + 1, 1);
  FOR(i, 1, n + 1) p[i] = p[i - 1] * 3;
  vector<vector<bool> > dp(1 << n);
  REP(i, 1 << n) dp[i].assign(x / p[n - __builtin_popcount(i)] + 1, false);
  REP(i, n) {
    if (a[i] < dp[1 << i].size()) dp[1 << i][a[i]] = true;
  }
  FOR(i, 1, 1 << n) REP(j, n) {
    if (i >> j & 1) continue;
    int nx_i = i | (1 << j);
    vector<bool> tmp(dp[nx_i].size(), false);
    REP(k, dp[i].size()) {
      if (dp[i][k]) {
        long long par = (1LL + k) * (a[j] + 1) - 1;
        if (par < tmp.size()) tmp[par] = true;
      }
    }
    REP(k, tmp.size()) {
      if (tmp[k]) {
        dp[nx_i][k] = true;
        if (k + a[j] < tmp.size()) tmp[k + a[j]] = true;
      }
    }
  }
  FOR(j, 1, x + 1) ans[j - 1] = dp[(1 << n) - 1][j];
  output(ans);
}