#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i)) #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i)) #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) #if defined(_MSC_VER) || __cplusplus > 199711L #define aut(r,v) auto r = (v) #else #define aut(r,v) __typeof(v) r = (v) #endif #define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it) #define all(o) (o).begin(), (o).end() #define pb(x) push_back(x) #define mp(x,y) make_pair((x),(y)) #define mset(m,v) memset(m,v,sizeof(m)) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3fLL using namespace std; typedef vector vi; typedef pair pii; typedef vector > vpii; typedef long long ll; template inline void amin(T &x, U y) { if(y < x) x = y; } template inline void amax(T &x, U y) { if(x < y) x = y; } int main() { int N; while(~scanf("%d", &N)) { vector c(N); rep(i, N) cin >> c[i]; sort(all(c)); vector dpmin(1 << N, INFL), dpmax(1 << N, -INFL); vector dpmin1(1 << N, INFL), dpmax1(1 << N, -INFL); reu(i, 1, 1 << N) { rep(k, N) if((i >> k & 1) && c[k][0] == '+') { int rem = i - (1 << k); for(int j = rem; j > 0; (-- j) &= rem) { amin(dpmin[i], dpmin[j] + dpmin[rem - j]); amax(dpmax[i], dpmax[j] + dpmax[rem - j]); } } rep(k, N) if((i >> k & 1) && c[k][0] == '-') { int rem = i - (1 << k); for(int j = rem; j > 0; (-- j) &= rem) { amin(dpmin[i], dpmin[j] - dpmax1[rem - j]); amax(dpmax[i], dpmax[j] - dpmin1[rem - j]); } } int ds = 0; ll num = 0, num2 = 0; rep(k, N) if((i >> k & 1) && isdigit(c[k][0])) { ds |= 1 << k; num = num * 10 + (c[k][0] - '0'); } for(int k = N - 1; k >= 0; -- k) if((i >> k & 1) && isdigit(c[k][0])) { num2 = num2 * 10 + (c[k][0] - '0'); } if(ds == i) { amin(dpmin[i], num); amax(dpmax[i], num); amin(dpmin[i], num2); amax(dpmax[i], num2); amin(dpmin1[i], num); amax(dpmax1[i], num); amin(dpmin1[i], num2); amax(dpmax1[i], num2); } // cerr << i << ": " << dpmax[i] << endl; } printf("%lld %lld\n", dpmax[(1 << N) - 1], dpmin[(1 << N) - 1]); } return 0; }