Problem:
Score : 400 points

Problem Statement

You are given a string S of length N consisting of ( and ). Your task is to insert some number of ( and ) into S to obtain a correct bracket sequence.

Here, a correct bracket sequence is defined as follows:

() is a correct bracket sequence.

If X is a correct bracket sequence, the concatenation of (, X and ) in this order is also a correct bracket sequence.

If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence.

Every correct bracket sequence can be derived from the rules above.

Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

Constraints

The length of S is N.

1 ≤ N ≤ 100

S consists of ( and ).

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the lexicographically smallest string among the shortest correct bracket sequences that can be obtained by inserting some number of ( and ) into S.

Sample Input 1

3
())

Sample Output 1

(())

Sample Input 2

6
)))())

Sample Output 2

(((()))())

Sample Input 3

8
))))((((

Sample Output 3

(((())))(((())))

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i, a, b) for(ll i = (a); i < (ll)b; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF (1ll << 60)
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll, ll> p;
constexpr int MOD = 1000000007;
using ll = long long;
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

void print(const std::vector<int> &v) {
    std::for_each(v.begin(), v.end(), [](int x) { std::cout << x << " "; });
    std::cout << std::endl;
}

int main() {
    int N;
    string S;
    cin >> N;
    cin >> S;
    int need_left = 0;
    int pointer = 0;
    for(int i = 0; i < S.size(); i++) {
        if(S[i] == '(') {
            pointer++;
        } else {
            //)のとき
            if(pointer == 0) {
                need_left++;
            } else {
                pointer--;
            }
        }
    }
    // pointerは残った(の数
    int need_right = pointer;
    string res = " ";
    for(int i = 0; i < need_left; i++) {
        res += "(";
    }
    res += S;
    for(int i = 0; i < need_right; i++) {
        res += ')';
    }
    cout << res << endl;
    return 0;
}