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.
)(()()))((
 )())))(((())))())))))))()())(())))))))())))(((())))())()()))(())())((()())(((((()()()())()()()))((()
 ()
 (((())))))()(()
 ))(
 )
 (()()
 )(
 (

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

//package vk2015.r1;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class A {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		char[] s = ns().toCharArray();
		int n = s.length;
		int[] h = new int[2*n+1];
		for(int i = 0;i < 2*n;i++){
			h[i+1] = h[i] + (s[i%n] == '(' ? 1 : -1);
		}
		SegmentTreeRMQ st = new SegmentTreeRMQ(h);
		
		char[] t = new char[2*n];
		for(int i = 0;i < n;i++){
			t[i+n] = t[i] = s[i];
		}
		int[] sa = sa(t);
		tr(h);
		tr(sa);
		for(int i = 0;i < 2*n;i++){
			if(sa[i] < n){
				tr(sa[i], st.min(sa[i], sa[i]+n+1), h[sa[i]+n]);
			}
			if(sa[i] < n && (st.min(sa[i], sa[i]+n+1) == h[sa[i]+n] || st.min(sa[i], sa[i]+n+1) == h[sa[i]])){
				for(int j = 0;j < -h[n];j++){
					out.print("(");
				}
				out.print(new String(t, sa[i], n));
				for(int j = 0;j < h[n];j++){
					out.print(")");
				}
				out.println();
				return;
			}
		}
	}
	
	
	private static interface BaseArray {
		public int get(int i);

		public void set(int i, int val);

		public int update(int i, int val);
	}

	private static class CharArray implements BaseArray {
		private char[] m_A = null;
		private int m_pos = 0;

		CharArray(char[] A, int pos) {
			m_A = A;
			m_pos = pos;
		}

		public int get(int i) {
			return m_A[m_pos + i] & 0xffff;
		}

		public void set(int i, int val) {
			m_A[m_pos + i] = (char) (val & 0xffff);
		}

		public int update(int i, int val) {
			return m_A[m_pos + i] += val & 0xffff;
		}
	}
	
	public static class SegmentTreeRMQ {
		public int M, H, N;
		public int[] st;
		
		public SegmentTreeRMQ(int n)
		{
			N = n;
			M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
			H = M>>>1;
			st = new int[M];
			Arrays.fill(st, 0, M, Integer.MAX_VALUE);
		}
		
		public SegmentTreeRMQ(int[] a)
		{
			N = a.length;
			M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
			H = M>>>1;
			st = new int[M];
			for(int i = 0;i < N;i++){
				st[H+i] = a[i];
			}
			Arrays.fill(st, H+N, M, Integer.MAX_VALUE);
			for(int i = H-1;i >= 1;i--)propagate(i);
		}
		
		public void update(int pos, int x)
		{
			st[H+pos] = x;
			for(int i = (H+pos)>>>1;i >= 1;i >>>= 1)propagate(i);
		}
		
		private void propagate(int i)
		{
			st[i] = Math.min(st[2*i], st[2*i+1]);
		}
		
		public int min(int l, int r){ return l >= r ? 0 : min(l, r, 0, H, 1);}
		
		private int min(int l, int r, int cl, int cr, int cur)
		{
			if(l <= cl && cr <= r){
				return st[cur];
			}else{
				int mid = cl+cr>>>1;
				int ret = Integer.MAX_VALUE;
				if(cl < r && l < mid){
					ret = Math.min(ret, min(l, r, cl, mid, 2*cur));
				}
				if(mid < r && l < cr){
					ret = Math.min(ret, min(l, r, mid, cr, 2*cur+1));
				}
				return ret;
			}
		}
		
		public int firstle(int l, int v) {
			int cur = H+l;
			while(true){
				if(st[cur] <= v){
					if(cur < H){
						cur = 2*cur;
					}else{
						return cur-H;
					}
				}else{
					cur++;
					if((cur&cur-1) == 0)return -1;
					if((cur&1)==0)cur>>>=1;
				}
			}
		}
		
		public int lastle(int l, int v) {
			int cur = H+l;
			while(true){
				if(st[cur] <= v){
					if(cur < H){
						cur = 2*cur+1;
					}else{
						return cur-H;
					}
				}else{
					if((cur&cur-1) == 0)return -1;
					cur--;
					if((cur&1)==1)cur>>>=1;
				}
			}
		}
	}

	private static class IntArray implements BaseArray {
		private int[] m_A = null;
		private int m_pos = 0;

		IntArray(int[] A, int pos) {
			m_A = A;
			m_pos = pos;
		}

		public int get(int i) {
			return m_A[m_pos + i];
		}

		public void set(int i, int val) {
			m_A[m_pos + i] = val;
		}

		public int update(int i, int val) {
			return m_A[m_pos + i] += val;
		}
	}

	/* find the start or end of each bucket */
	private static void getCounts(BaseArray T, BaseArray C, int n, int k) {
		int i;
		for(i = 0;i < k;++i){
			C.set(i, 0);
		}
		for(i = 0;i < n;++i){
			C.update(T.get(i), 1);
		}
	}

	private static void getBuckets(BaseArray C, BaseArray B, int k, boolean end) {
		int i, sum = 0;
		if(end != false){
			for(i = 0;i < k;++i){
				sum += C.get(i);
				B.set(i, sum);
			}
		}else{
			for(i = 0;i < k;++i){
				sum += C.get(i);
				B.set(i, sum - C.get(i));
			}
		}
	}

	/* sort all type LMS suffixes */
	private static void LMSsort(BaseArray T, int[] SA, BaseArray C,
			BaseArray B, int n, int k) {
		int b, i, j;
		int c0, c1;
		/* compute SAl */
		if(C == B){
			getCounts(T, C, n, k);
		}
		getBuckets(C, B, k, false); /* find starts of buckets */
		j = n - 1;
		b = B.get(c1 = T.get(j));
		--j;
		SA[b++] = (T.get(j) < c1) ? ~j : j;
		for(i = 0;i < n;++i){
			if(0 < (j = SA[i])){
				if((c0 = T.get(j)) != c1){
					B.set(c1, b);
					b = B.get(c1 = c0);
				}
				--j;
				SA[b++] = (T.get(j) < c1) ? ~j : j;
				SA[i] = 0;
			}else if(j < 0){
				SA[i] = ~j;
			}
		}
		/* compute SAs */
		if(C == B){
			getCounts(T, C, n, k);
		}
		getBuckets(C, B, k, true); /* find ends of buckets */
		for(i = n - 1, b = B.get(c1 = 0);0 <= i;--i){
			if(0 < (j = SA[i])){
				if((c0 = T.get(j)) != c1){
					B.set(c1, b);
					b = B.get(c1 = c0);
				}
				--j;
				SA[--b] = (T.get(j) > c1) ? ~(j + 1) : j;
				SA[i] = 0;
			}
		}
	}

	private static int LMSpostproc(BaseArray T, int[] SA, int n, int m) {
		int i, j, p, q, plen, qlen, name;
		int c0, c1;
		boolean diff;

		/*
		 * compact all the sorted substrings into the first m items of SA 2*m
		 * must be not larger than n (proveable)
		 */
		for(i = 0;(p = SA[i]) < 0;++i){
			SA[i] = ~p;
		}
		if(i < m){
			for(j = i, ++i;;++i){
				if((p = SA[i]) < 0){
					SA[j++] = ~p;
					SA[i] = 0;
					if(j == m){
						break;
					}
				}
			}
		}

		/* store the length of all substrings */
		i = n - 1;
		j = n - 1;
		c0 = T.get(n - 1);
		do{
			c1 = c0;
		}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
		for(;0 <= i;){
			do{
				c1 = c0;
			}while ((0 <= --i) && ((c0 = T.get(i)) <= c1));
			if(0 <= i){
				SA[m + ((i + 1) >> 1)] = j - i;
				j = i + 1;
				do{
					c1 = c0;
				}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
			}
		}

		/* find the lexicographic names of all substrings */
		for(i = 0, name = 0, q = n, qlen = 0;i < m;++i){
			p = SA[i];
			plen = SA[m + (p >> 1)];
			diff = true;
			if((plen == qlen) && ((q + plen) < n)){
				for(j = 0;(j < plen) && (T.get(p + j) == T.get(q + j));++j){
				}
				if(j == plen){
					diff = false;
				}
			}
			if(diff != false){
				++name;
				q = p;
				qlen = plen;
			}
			SA[m + (p >> 1)] = name;
		}

		return name;
	}

	/* compute SA and BWT */
	private static void induceSA(BaseArray T, int[] SA, BaseArray C,
			BaseArray B, int n, int k) {
		int b, i, j;
		int c0, c1;
		/* compute SAl */
		if(C == B){
			getCounts(T, C, n, k);
		}
		getBuckets(C, B, k, false); /* find starts of buckets */
		j = n - 1;
		b = B.get(c1 = T.get(j));
		SA[b++] = ((0 < j) && (T.get(j - 1) < c1)) ? ~j : j;
		for(i = 0;i < n;++i){
			j = SA[i];
			SA[i] = ~j;
			if(0 < j){
				if((c0 = T.get(--j)) != c1){
					B.set(c1, b);
					b = B.get(c1 = c0);
				}
				SA[b++] = ((0 < j) && (T.get(j - 1) < c1)) ? ~j : j;
			}
		}
		/* compute SAs */
		if(C == B){
			getCounts(T, C, n, k);
		}
		getBuckets(C, B, k, true); /* find ends of buckets */
		for(i = n - 1, b = B.get(c1 = 0);0 <= i;--i){
			if(0 < (j = SA[i])){
				if((c0 = T.get(--j)) != c1){
					B.set(c1, b);
					b = B.get(c1 = c0);
				}
				SA[--b] = ((j == 0) || (T.get(j - 1) > c1)) ? ~j : j;
			}else{
				SA[i] = ~j;
			}
		}
	}

	/*
	 * find the suffix array SA of T[0..n-1] in {0..k-1}^n use a working space
	 * (excluding T and SA) of at most 2n+O(1) for a constant alphabet
	 */
	private static void SA_IS(BaseArray T, int[] SA, int fs, int n, int k) {
		BaseArray C, B, RA;
		int i, j, b, m, p, q, name, newfs;
		int c0, c1;
		int flags = 0;

		if(k <= 256){
			C = new IntArray(new int[k], 0);
			if(k <= fs){
				B = new IntArray(SA, n + fs - k);
				flags = 1;
			}else{
				B = new IntArray(new int[k], 0);
				flags = 3;
			}
		}else if(k <= fs){
			C = new IntArray(SA, n + fs - k);
			if(k <= (fs - k)){
				B = new IntArray(SA, n + fs - k * 2);
				flags = 0;
			}else if(k <= 1024){
				B = new IntArray(new int[k], 0);
				flags = 2;
			}else{
				B = C;
				flags = 8;
			}
		}else{
			C = B = new IntArray(new int[k], 0);
			flags = 4 | 8;
		}

		/*
		 * stage 1: reduce the problem by at least 1/2 sort all the
		 * LMS-substrings
		 */
		getCounts(T, C, n, k);
		getBuckets(C, B, k, true); /* find ends of buckets */
		for(i = 0;i < n;++i){
			SA[i] = 0;
		}
		b = -1;
		i = n - 1;
		j = n;
		m = 0;
		c0 = T.get(n - 1);
		do{
			c1 = c0;
		}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
		for(;0 <= i;){
			do{
				c1 = c0;
			}while ((0 <= --i) && ((c0 = T.get(i)) <= c1));
			if(0 <= i){
				if(0 <= b){
					SA[b] = j;
				}
				b = B.update(c1, -1);
				j = i;
				++m;
				do{
					c1 = c0;
				}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
			}
		}
		if(1 < m){
			LMSsort(T, SA, C, B, n, k);
			name = LMSpostproc(T, SA, n, m);
		}else if(m == 1){
			SA[b] = j + 1;
			name = 1;
		}else{
			name = 0;
		}

		/*
		 * stage 2: solve the reduced problem recurse if names are not yet
		 * unique
		 */
		if(name < m){
			if((flags & 4) != 0){
				C = null;
				B = null;
			}
			if((flags & 2) != 0){
				B = null;
			}
			newfs = (n + fs) - (m * 2);
			if((flags & (1 | 4 | 8)) == 0){
				if((k + name) <= newfs){
					newfs -= k;
				}else{
					flags |= 8;
				}
			}
			for(i = m + (n >> 1) - 1, j = m * 2 + newfs - 1;m <= i;--i){
				if(SA[i] != 0){
					SA[j--] = SA[i] - 1;
				}
			}
			RA = new IntArray(SA, m + newfs);
			SA_IS(RA, SA, newfs, m, name);
			RA = null;

			i = n - 1;
			j = m * 2 - 1;
			c0 = T.get(n - 1);
			do{
				c1 = c0;
			}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
			for(;0 <= i;){
				do{
					c1 = c0;
				}while ((0 <= --i) && ((c0 = T.get(i)) <= c1));
				if(0 <= i){
					SA[j--] = i + 1;
					do{
						c1 = c0;
					}while ((0 <= --i) && ((c0 = T.get(i)) >= c1));
				}
			}

			for(i = 0;i < m;++i){
				SA[i] = SA[m + SA[i]];
			}
			if((flags & 4) != 0){
				C = B = new IntArray(new int[k], 0);
			}
			if((flags & 2) != 0){
				B = new IntArray(new int[k], 0);
			}
		}

		/* stage 3: induce the result for the original problem */
		if((flags & 8) != 0){
			getCounts(T, C, n, k);
		}
		/* put all left-most S characters into their buckets */
		if(1 < m){
			getBuckets(C, B, k, true); /* find ends of buckets */
			i = m - 1;
			j = n;
			p = SA[m - 1];
			c1 = T.get(p);
			do{
				q = B.get(c0 = c1);
				while (q < j){
					SA[--j] = 0;
				}
				do{
					SA[--j] = p;
					if(--i < 0){
						break;
					}
					p = SA[i];
				}while ((c1 = T.get(p)) == c0);
			}while (0 <= i);
			while (0 < j){
				SA[--j] = 0;
			}
		}
		induceSA(T, SA, C, B, n, k);
		C = null;
		B = null;
	}

	/* char */
	public static void suffixsort(char[] T, int[] SA, int n) {
		if((T == null) || (SA == null) || (T.length < n) || (SA.length < n)){
			return;
		}
		if(n <= 1){
			if(n == 1){
				SA[0] = 0;
			}
			return;
		}
		SA_IS(new CharArray(T, 0), SA, 0, n, 128);
	}
	
	public static int[] sa(char[] T)
	{
		int n = T.length;
		int[] sa = new int[n];
		suffixsort(T, sa, n);
		return sa;
	}
	
	public static int[] buildLCP(char[] str, int[] sa)
	{
		int n = str.length;
		int h = 0;
		int[] lcp = new int[n];
		int[] isa = new int[n];
		for(int i = 0;i < n;i++)isa[sa[i]] = i;
		for(int i = 0;i < n;i++){
			if(isa[i] > 0){
				for(int j = sa[isa[i]-1]; j+h<n && i+h<n && str[j+h] == str[i+h]; h++);
				lcp[isa[i]] = h;
			}else{
				lcp[isa[i]] = 0;
			}
			if(h > 0)h--;
		}
		return lcp;
	}
	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new A().run(); }
	
	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
