cf149DColoring Brackets【区间dp求方法数】

Description

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.

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).

Sample Input

Input

(())

Output

Input

(()())

Output

Input

()

Output

Hint

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.

The two ways of coloring shown below are incorrect.

rt,这个题是求方法数的==

题意：为匹配好的括号涂色，满足三个条件：1）要么不涂色，要么涂红色，要么涂蓝色 2）每对匹配的括号有且只有一个涂色 3）相邻的两个括号不同色

一说到括号匹配应该马上想到用栈来找匹配的括号并记录序号对应关系，找好之后，dp无外乎是递推或是递归，递推一般用来解决“序列每个元素等级相同”的问题，就比方说那种第一重循环是区间长度，第二重循环是区间左端点的，明显和这个题对不上。递归的话，最开始想的是找到一段(((..)))这种的就调用一次，其实这个步骤完全可以写在dfs中，而且之前的想法不能够求((..)(..))这种，难道求的时候还要继续用栈？当然不是！按照正常人的思路第一层dfs一定是(0,n-1)

如果区间左右端点是匹配的，皆大欢喜，我们dfs中间部分；如果不匹配，那么我们把他分成匹配的两段递归就好了啊~dp的转移方程写着不难，可以合并分情况讨论，我居然是#define mod 1000000007出问题了 const是自带括号的，define写啥就是啥！！不能写成#define mod 1000000000+7！！

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
#define mod 1000000007
char str[1000];
int num[1000];
long long dp[1000][1000][3][3];
int n;
void dfs(int l,int r)
{
    if(l+1==r)
    {
        dp[l][r][1][0]=dp[l][r][0][1]=dp[l][r][0][2]=dp[l][r][2][0]=1;
        return;
    }
    if(r==num[l])
    {
        dfs(l+1,r-1);
        for(int i=0;i<3;i++)
            for(int j=0;j<3;j++)
            {
                if(i!=1)dp[l][r][1][0]=(dp[l][r][1][0]+dp[l+1][r-1][i][j])%mod;
                if(j!=1)dp[l][r][0][1]=(dp[l][r][0][1]+dp[l+1][r-1][i][j])%mod;
                if(i!=2)dp[l][r][2][0]=(dp[l][r][2][0]+dp[l+1][r-1][i][j])%mod;
                if(j!=2)dp[l][r][0][2]=(dp[l][r][0][2]+dp[l+1][r-1][i][j])%mod;
            }
        return;
    }
    int t=num[l];
    dfs(l,t);
    dfs(t+1,r);
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
            for(int k=0;k<3;k++)
                for(int q=0;q<3;q++)
                    if(!((k==1&&q==1)||(k==2&&q==2)))
                        dp[l][r][i][j]=(dp[l][r][i][j]+(dp[l][t][i][k]*dp[t+1][r][q][j])%mod)%mod;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    while(~scanf("%s",str))
    {
        n=strlen(str);
        stack<int>s;
        for(int i=0;i<n;i++)
        {
            if(str[i]=='(')s.push(i);
            else
            {
                int t=s.top();
                num[t]=i;
                num[i]=t;
                s.pop();
            }
        }
       // for(int i=0;i<n;i++)printf("i=%d,num=%d\n",i,num[i]);
        memset(dp,0,sizeof(dp));
        dfs(0,n-1);
        long long ans=0;
        for(int i=0;i<3;i++)
            for(int j=0;j<3;j++)
                ans=(ans+dp[0][n-1][i][j])%mod;
        printf("%I64d\n",ans);
    }
    return 0;
}