#include #include #include #include //#include #include #include #include #include #include //#include #include #include #include //#include #include #include //#include #include #include #include #include const int dx[] = {1, 0, -1, 0}; const int dy[] = {0, 1, 0, -1}; using namespace std; typedef long long ll; typedef vector vi; typedef vector vll; typedef pair pii; const int MAXN = 111; const int MAX = 50000; pii A[MAXN]; int dp[MAXN*MAX]; bool use[MAXN]; int main() { cin.tie(0); ios::sync_with_stdio(false); int N; ll X; cin >> N >> X; for (int i = 0; i < N; i++) { cin >> A[i].first; A[i].second = i; } sort(A, A+N); int cnt = 0; memset(dp, -1, sizeof(dp)); dp[0] = 0; for (; cnt < N && A[cnt].first < MAX; cnt++) {} for (int i = 0; i < cnt; i++) { for (int j = i*A[i].first; j >= 0; j--) { if (dp[j] == -1) continue; dp[j+A[i].first] = i+1; } } int rest = N-cnt; for (int s = 0; s < 1<>i)&1) sum += A[cnt+i].first; } ll tar = X-sum; if (0 <= tar && tar < MAXN*MAX && dp[tar] != -1) { for (int i = 0; i < rest; i++) { if ((s>>i)&1) use[i+cnt] = true; } int now = tar; while (now != 0) { use[A[dp[now]-1].second] = true; now -= A[dp[now]-1].first; } for (int i = 0; i < N; i++) { if (use[i]) cout << "o"; else cout << "x"; } cout << endl; return 0; } } cout << "No" << endl; return 0; }