（CodeForces

（CodeForces - 149D）Coloring Brackets

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

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).

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.

The two ways of coloring shown below are incorrect.

题目大意：给出一个合法的括号序列，给这个括号序列上色，要求：

• 对于每一个括号只有三种选择，不上色，上红色，上蓝色
• 相邻两个括号不能上同一种颜色
• 两个相互匹配的括号能且只能一个上色

先给出有n个合法的括号序列问有几种上色方式

思路：区间dp。设f[l][r][i][j]表示区间[l,r]的括号序列，两端颜色分别是i，j的上色方式（其中0代表不上色，1代表上红色，2代表上蓝色）。先预处理出每一个括号与它匹配的另一个括号的位置，即match[i]表示与位置为i的括号匹配的那个括号的位置。因为有四维所以采用记忆化搜索的方式写比较好，一比较好懂，现在就有如下几种情况：
1、如果(l+1==r) 说明就只有一对括号，显然可得
f[l][r][0][1]=1
f[l][r][1][0]=1
f[l][r][0][2]=1
f[l][r][2][0]=1
2、如果(l与r是配对的)
那么就可以递归下去，即dfs（l+1，r-1）
而区间[l,r]由于两端的括号能且只能一个上色，所以可得以下四种状态转移方程：
f[l][r][0][1]=(f[l][r][0][1]+f[l+1][r−1][i][j])%mod
f[l][r][1][0]=(f[l][r][1][0]+f[l+1][r−1][i][j])%mod
f[l][r][0][2]=(f[l][r][0][2]+f[l+1][r−1][i][j])%mod
f[l][r][2][0]=(f[l][r][2][0]+f[l+1][r−1][i][j])%mod
3、如果(l与r不配对)，那么可以枚举中间的匹配点k=match[i]，那么根据上色要求可以得到状态转移方程是：
f[l][r][i][j]=(f[l][r][i][j]+(f[l][k][i][m]∗f[k+1][r][n][j])%mod)%mod
其中因为k，k+1相邻所以颜色m，n不能相等又0<=m，n<=1

#include<cstdio>#include<cstring>#include<stack>using namespace std;typedef long long LL;const int maxn=705;const int mod=1e9+7;char a[maxn];int match[maxn];LL f[maxn][maxn][3][3];//0不涂色，1红色，2蓝色 void getmatch(int len){    stack<int> st;    for(int i=0;i<len;i++)    {        if(a[i]=='(') st.push(i);        else        {            match[i]=st.top();            match[st.top()]=i;            st.pop();        }    }}void dfs(int l,int r){    if(l+1==r)    {        f[l][r][0][1]=1;        f[l][r][1][0]=1;         f[l][r][0][2]=1;        f[l][r][2][0]=1;        return;    }    if(match[l]==r)    {        dfs(l+1,r-1);        for(int i=0;i<3;i++)            for(int j=0;j<3;j++)            {                if(j!=1)                    f[l][r][0][1]=(f[l][r][0][1]+f[l+1][r-1][i][j])%mod;                if(i!=1)                     f[l][r][1][0]=(f[l][r][1][0]+f[l+1][r-1][i][j])%mod;                if(j!=2)                    f[l][r][0][2]=(f[l][r][0][2]+f[l+1][r-1][i][j])%mod;                if(i!=2)                    f[l][r][2][0]=(f[l][r][2][0]+f[l+1][r-1][i][j])%mod;            }        return;    }    else    {        int k=match[l];        dfs(l,k);        dfs(k+1,r);        for(int i=0;i<3;i++)            for(int j=0;j<3;j++)                for(int m=0;m<3;m++)                    for(int n=0;n<3;n++)                        if(!((m==1&&n==1)||(m==2&&n==2)))                            f[l][r][i][j]=(f[l][r][i][j]+(f[l][k][i][m]*f[k+1][r][n][j])%mod)%mod;    }}int main(){    while(~scanf("%s",a))    {        int len=strlen(a);        memset(f,0,sizeof(f));        getmatch(len);        dfs(0,len-1);        LL ans=0;        for(int i=0;i<3;i++)            for(int j=0;j<3;j++)                ans=(ans+f[0][len-1][i][j])%mod;        printf("%lld\n",ans);    }    return 0;}