Problem: Polycarpus has a finite sequence of opening and closing brackets. In order not to fall asleep in a lecture, Polycarpus is having fun with his sequence. He is able to perform two operations:

  * adding any bracket in any position (in the beginning, the end, or between any two existing brackets); 
  * cyclic shift — moving the last bracket from the end of the sequence to the beginning. 



Polycarpus can apply any number of operations to his sequence and adding a cyclic shift in any order. As a result, he wants to get the correct bracket sequence of the minimum possible length. If there are several such sequences, Polycarpus is interested in the lexicographically smallest one. Help him find such a sequence.

Acorrect bracket sequence is a sequence of opening and closing brackets, from which you can get a correct arithmetic expression by adding characters "1" and "+" . Each opening bracket must correspond to a closed one. For example, the sequences "(())()", "()", "(()(()))" are correct and ")(", "(()" and "(()))(" are not.

The sequence a1 a2... an is lexicographically smaller than sequence b1 b2... bn, if there is such number i from 1 to n, thatak = bk for 1 ≤ k < i and ai < bi. Consider that "("  < ")".

Input

The first line contains Polycarpus's sequence consisting of characters "(" and ")". The length of a line is from 1 to 1 000 000.

Output

Print a correct bracket sequence of the minimum length that Polycarpus can obtain by his operations. If there are multiple such sequences, print the lexicographically minimum one.

Examples

Input

()(())


Output

(())()

Input

()(


Output

(())

Note

The sequence in the first example is already correct, but to get the lexicographically minimum answer, you need to perform four cyclic shift operations. In the second example you need to add a closing parenthesis between the second and third brackets and make a cyclic shift. You can first make the shift, and then add the bracket at the end.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class A, class B, class C>
struct Triple {
  A first;
  B second;
  C third;
};
template <class T>
void ResizeVec(T&, vector<long long>) {}
template <class T>
void ResizeVec(vector<T>& vec, vector<long long> sz) {
  vec.resize(sz[0]);
  sz.erase(sz.begin());
  if (sz.empty()) {
    return;
  }
  for (T& v : vec) {
    ResizeVec(v, sz);
  }
}
inline bool leq(long long a1, long long a2, long long b1, long long b2) {
  return (a1 < b1 || a1 == b1 && a2 <= b2);
}
inline bool leq(long long a1, long long a2, long long a3, long long b1,
                long long b2, long long b3) {
  return (a1 < b1 || a1 == b1 && leq(a2, a3, b2, b3));
}
static void radixPass(long long* a, long long* b, long long* r, long long n,
                      long long K) {
  long long* c = new long long[K + 1];
  for (long long i = 0; i <= K; i++) c[i] = 0;
  for (long long i = 0; i < n; i++) c[r[a[i]]]++;
  for (long long i = 0, sum = 0; i <= K; i++) {
    long long t = c[i];
    c[i] = sum;
    sum += t;
  }
  for (long long i = 0; i < n; i++) b[c[r[a[i]]]++] = a[i];
  delete[] c;
}
void suffixArray(long long* s, long long* SA, long long n, long long K) {
  long long n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
  long long* s12 = new long long[n02 + 3];
  s12[n02] = s12[n02 + 1] = s12[n02 + 2] = 0;
  long long* SA12 = new long long[n02 + 3];
  SA12[n02] = SA12[n02 + 1] = SA12[n02 + 2] = 0;
  long long* s0 = new long long[n0];
  long long* SA0 = new long long[n0];
  for (long long i = 0, j = 0; i < n + (n0 - n1); i++)
    if (i % 3 != 0) s12[j++] = i;
  radixPass(s12, SA12, s + 2, n02, K);
  radixPass(SA12, s12, s + 1, n02, K);
  radixPass(s12, SA12, s, n02, K);
  long long name = 0, c0 = -1, c1 = -1, c2 = -1;
  for (long long i = 0; i < n02; i++) {
    if (s[SA12[i]] != c0 || s[SA12[i] + 1] != c1 || s[SA12[i] + 2] != c2) {
      name++;
      c0 = s[SA12[i]];
      c1 = s[SA12[i] + 1];
      c2 = s[SA12[i] + 2];
    }
    if (SA12[i] % 3 == 1)
      s12[SA12[i] / 3] = name;
    else
      s12[SA12[i] / 3 + n0] = name;
  }
  if (name < n02) {
    suffixArray(s12, SA12, n02, name);
    for (long long i = 0; i < n02; i++) s12[SA12[i]] = i + 1;
  } else
    for (long long i = 0; i < n02; i++) SA12[s12[i] - 1] = i;
  for (long long i = 0, j = 0; i < n02; i++)
    if (SA12[i] < n0) s0[j++] = 3 * SA12[i];
  radixPass(s0, SA0, s, n0, K);
  for (long long p = 0, t = n0 - n1, k = 0; k < n; k++) {
    long long i = (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2);
    long long j = SA0[p];
    if (SA12[t] < n0 ? leq(s[i], s12[SA12[t] + n0], s[j], s12[j / 3])
                     : leq(s[i], s[i + 1], s12[SA12[t] - n0 + 1], s[j],
                           s[j + 1], s12[j / 3 + n0])) {
      SA[k] = i;
      t++;
      if (t == n02)
        for (k++; p < n0; p++, k++) SA[k] = SA0[p];
    } else {
      SA[k] = j;
      p++;
      if (p == n0)
        for (k++; t < n02; t++, k++)
          SA[k] = (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2);
    }
  }
  delete[] s12;
  delete[] SA12;
  delete[] SA0;
  delete[] s0;
}
const long long N = 2e6 + 5;
long long s[N];
long long SA[N];
vector<long long> MinK(vector<long long> vec, long long k) {
  vector<long long> res;
  long long n = ((long long)(vec).size());
  if (n - k + 1 <= 0) {
    return vector<long long>();
  }
  res.resize(n - k + 1);
  vector<long long> que;
  long long beg;
  for (long long i = 0; i < ((long long)(vec).size()); i++) {
    while (!que.empty() && vec[que.back()] >= vec[i]) {
      que.pop_back();
    }
    que.push_back(i);
    beg = min(beg, ((long long)(que).size()) - 1);
    if (que[beg] <= i - k) {
      beg++;
    }
    if (i >= k - 1) {
      res[i - k + 1] = vec[que[beg]];
    }
  }
  for (long long i = (1); i <= (k - 1); ++i) {
    res.push_back(N);
  }
  return res;
}
long long SAinv[N];
int main() {
  ios_base::sync_with_stdio(0);
  string br;
  cin >> br;
  ;
  long long n = ((long long)(br).size());
  br = ' ' + br;
  for (long long i = (1); i <= (n); ++i) {
    if (br[i] == '(') {
      s[i] = 1;
    } else {
      s[i] = 2;
    }
  }
  vector<long long> bilv(2 * n + 5);
  bilv.push_back(0);
  for (long long i = (n + 1); i <= (2 * n); ++i) {
    s[i] = s[i - n];
    br += br[i - n];
  }
  long long b = 0;
  for (long long i = (1); i <= (2 * n); ++i) {
    if (s[i] == 1) {
      b++;
    } else {
      b--;
    }
    bilv[i] = b;
  }
  long long bil = bilv[n];
  suffixArray(s + 1, SA + 1, 2 * n, 2);
  for (long long i = (1); i <= (2 * n); ++i) {
    SAinv[SA[i] + 1] = i;
  }
  vector<long long> mins = MinK(bilv, n);
  ;
  ;
  ;
  long long best = 0;
  SAinv[0] = N;
  for (long long i = (0); i <= (n - 1); ++i) {
    if (mins[i + 1] >= bilv[i] + min(0ll, bil)) {
      if (SAinv[i + 1] < SAinv[best]) {
        ;
        best = i + 1;
      }
    }
  }
  if (bil < 0) {
    for (long long i = (1); i <= (-bil); ++i) {
      putchar('(');
    }
  }
  for (long long i = (best); i <= (best + n - 1); ++i) {
    putchar(br[i]);
  }
  if (bil >= 0) {
    for (long long i = (1); i <= (bil); ++i) {
      putchar(')');
    }
  }
  return 0;
}