#include #include #include using namespace std; template inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; } template inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; } template inline void POSS(T condition){ if(condition) cout << "POSSIBLE" << endl; else cout << "IMPOSSIBLE" << endl; } template inline void Poss(T condition){ if(condition) cout << "Possible" << endl; else cout << "Impossible" << endl; } template inline void First(T condition){ if(condition) cout << "First" << endl; else cout << "Second" << endl; } int character_count(string text, char character){ int ans = 0; for(int i = 0; i < text.size(); i++){ ans += (text[i] == character); } return ans; } long power(long base, long exponent, long module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ long root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }} struct position{ int y, x; }; position move_pattern[4] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); } template string to_string(pair x){ return to_string(x.first) + "," + to_string(x.second); } template void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++){ ans += to_string(*i) + " "; } ans.pop_back(); cout << ans << endl; } template T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template T lcm(T a, T b){ return a / gcd(a, b) * b; } #define mod long(1e9 + 7) #define all(x) (x).begin(), (x).end() #define bitcount(n) __builtin_popcountl(long(n)) #define fcout cout << fixed << setprecision(10) #define highest(x) (63 - __builtin_clzl(x)) int main(){ int N; cin >> N; int ans1 = 1e9; int a[N]; cin >> a[0]; for(int i = 1; i < N; i++){ cin >> a[i]; ans1 += min(ans1, a[i] - a[i - 1]); } cout << ans1 << " " << a[N - 1] - a[0] << endl; }