#include #define REP(i,n) for(int i=0;i<(int)(n);i++) #define ALL(x) (x).begin(),(x).end() using namespace std; typedef long long ll; int dp[4096000]; string solve() { int N; ll X; cin >> N >> X; memset(dp, -1, sizeof(dp)); dp[0] = 0; vector A(N); vector> B; REP(i,N) { cin >> A[i]; if (A[i] < 100000) { for (int j = 2010000; j >= 0; --j) if (dp[j] >= 0 && dp[j+A[i]] == -1) dp[j+A[i]] = i; } else B.emplace_back(i, A[i]); } int M = B.size(); REP(bit,1<> i) & 1) sum += B[i].second; if (0 <= X - sum && X - sum < 4096000 && dp[X-sum] >= 0) { string res(N, 'x'); REP(i,M) if ((bit >> i) & 1) res[B[i].first] = 'o'; for (int val = X - sum; val > 0; val -= A[dp[val]]) res[dp[val]] = 'o'; return res; } } return "No"; } int main() { cout << solve() << endl; }