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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <array>
#include <string>
#include <iomanip>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define mp make_pair
#define eb emplace_back
#define F first
#define S second
using namespace std;
using namespace __gnu_pbds;

typedef long long int ll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ost;

const ll maxN = 1e7;
const ll INF = 1e17;
const ll MOD = 1e9 + 7;

ll xs[] = {-1, 1, 0, 0, -1, 1, 1, -1};
ll ys[] = {0, 0, -1, 1, -1, 1, -1, 1};

ll segtree[maxN << 2];
ll lazy[maxN << 2];
ll A[maxN];

inline ll left(ll x){
	return x << 1;
}

inline ll right(ll x){
	return (x<<1) + 1;
}

void build(ll node, ll start, ll end){
	if(start == end)
		segtree[node] = A[start];
	else{
		ll mid = (start + end)/2;
		build(left(node), start, mid);
		build(right(node), mid+1, end);
		segtree[node] = segtree[left(node)] + segtree[right(node)];
	}
}

void update(ll node, ll start, ll end, ll l, ll r, ll val){
	if(lazy[node] != 0){
		segtree[node] += (end-start+1)*lazy[node];
		if(start != end){
			lazy[left(node)] += lazy[node];
			lazy[right(node)] += lazy[node];
		}
		lazy[node] = 0;
	}
	if(start > end or l > start or r < end)
		return ;
	else if(start >= l and r >= end){
		segtree[node] += val;
		if(start != end){
			lazy[left(node)] += val;
			lazy[right(node)] += val;
		}
	}
	else{
		ll mid=(start + end)/2;
		update(left(node), start, mid, l, r, val);
		update(right(node), mid+1, end, l, r, val);
		segtree[node] = segtree[left(node)] + segtree[right(node)];
	}
}

ll sums(ll node, ll start, ll end, ll l, ll r){
	if(lazy[node] != 0){
		segtree[node] += (end-start+1)*lazy[node];
		if(start != end){
			lazy[left(node)] += lazy[node];
			lazy[right(node)] += lazy[node];
		}
		lazy[node] = 0;
	}
	if(l > end or r < start or start > end)
		return 0;
	if(l <= start and r >= end)
		return segtree[node];
	ll mid = (start+end)/2;
	return sums(left(node), start, mid, l , r) + sums(right(node), mid + 1, end, l , r);
}

int main(){
	cin.sync_with_stdio(false);
	cin.tie(0);
	ll n;
	string s;
	cin >> n >> s;
	stack<char> stc;
	string ans = "";
	for(auto x : s){
		if(x == '('){
			stc.push(x);
			ans += x;
		}
		else{
			if(stc.empty())
				ans = '(' + ans + ')';
			else{
				stc.pop();
				ans += ')';
			}
		}
	}
	while(!stc.empty()){
		stc.pop();
		ans+=')';
	}
	cout << ans;
	return 0;
}