Problem: Once Petya read a problem about a bracket sequence. He gave it much thought but didn't find a solution. Today you will face it.

You are given string s. It represents a correct bracket sequence. A correct bracket sequence is the sequence of opening ("(") and closing (")") brackets, such that it is possible to obtain a correct mathematical expression from it, inserting numbers and operators between the brackets. For example, such sequences as "(())()" and "()" are correct bracket sequences and such sequences as ")()" and "(()" are not.

In a correct bracket sequence each bracket corresponds to the matching bracket (an opening bracket corresponds to the matching closing bracket and vice versa). For example, in a bracket sequence shown of the figure below, the third bracket corresponds to the matching sixth one and the fifth bracket corresponds to the fourth one.

<image>

You are allowed to color some brackets in the bracket sequence so as all three conditions are fulfilled: 

  * Each bracket is either not colored any color, or is colored red, or is colored blue. 
  * For any pair of matching brackets exactly one of them is colored. In other words, for any bracket the following is true: either it or the matching bracket that corresponds to it is colored. 
  * No two neighboring colored brackets have the same color. 



Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. Two ways of coloring are considered different if they differ in the color of at least one bracket. As the result can be quite large, print it modulo 1000000007 (109 + 7).

Input

The first line contains the single string s (2 ≤ |s| ≤ 700) which represents a correct bracket sequence. 

Output

Print the only number — the number of ways to color the bracket sequence that meet the above given conditions modulo 1000000007 (109 + 7).

Examples

Input

(())


Output

12


Input

(()())


Output

40


Input

()


Output

4

Note

Let's consider the first sample test. The bracket sequence from the sample can be colored, for example, as is shown on two figures below. 

<image> <image>

The two ways of coloring shown below are incorrect. 

<image> <image>

Solution (in JAVA):

Solution:
/*   合    */
import java.io.*;
import java.util.*;
import java.math.*;

public final class coloring_brackets
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static long[][][][] dp;
	static int maxn=720;
	static int[] arr;
	static long mod=(long)(1e9+7);
	
	static long solve(int l,int r,int lc,int rc) throws Exception
	{
		
		if(l>r)
		{
			return 1;
		}
		
		if(l==r)
		{
			System.out.println(l);
			
			throw new Exception("Impossible");
		}
		
		if(dp[l][r][lc][rc] != -1)
		{
			return dp[l][r][lc][rc];
		}
		
		else
		{
			long ret=0;int curr=arr[l];
			
			for(int i=0;i<=1;i++)
			{
				if(i != lc)
				{
					long now = (solve(l+1,curr-1,i,2)*solve(curr+1,r,2,rc))%mod;
					
					ret=(ret+now)%mod;
				}
				
				if((curr+1<=r) || i!=rc)
				{
					long now = (solve(l+1,curr-1,2,i)*solve(curr+1,r,i,rc))%mod;
				
					ret=(ret+now)%mod;
				}
			}
			
			dp[l][r][lc][rc]=ret; return ret;
		}
	}
		
    public static void main(String args[]) throws Exception
    {
		char[] a=sc.next().toCharArray(); dp=new long[maxn][maxn][3][3];arr=new int[a.length];
		
		Stack<Integer> st=new Stack<>();
		
		for(int i=0;i<a.length;i++)
		{
			if(a[i]=='(')
			{
				st.push(i);
			}
			else
			{
				arr[st.pop()]=i;
			}
		}
		
		for(int i=0;i<maxn;i++)
		{
			for(int j=0;j<maxn;j++)
			{
				for(int k=0;k<3;k++)
				{
					for(int x=0;x<3;x++)
					{
						dp[i][j][k][x]=-1;
					}
				}
			}
		}
		
		out.println(solve(0,a.length-1,2,2)); out.close();
		
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}