HZNU Training 3—L

L - Coloring Brackets

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).

Example

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.

 

【分析】

题意：给一组括号序列上色，规则如下：

1.只有三种颜色，无色，红色，蓝色

2.每对括号只能有一个上色

3.相邻的两个括号不能上同样的颜色，可以都无色

问给定的括号序列有多少种满足规则的方案总数

感觉很多题目都喜欢用括号做文章....很明显的区间dp吧

写成递归可能好理解一点...

//0—无色  1—红色 2—蓝色

f[l][r][i][j]表示区间[l,r]，l的颜色为i，r的颜色为j的方案总数

所以对当前状态考虑三个情况

1.若l和r为一对括号并且他们内部没有括号，那么表示递归到达底层初始化

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.若l和r为一对括号，那么往内搜索l-1,r-1的状态

搜索完后以f[l-1][r-1]更新当前四个状态f[l][r][0][1],f[l][r][0][2],f[l][r][1][0],f[l][r][2][0];

模拟一下f[l-1][r-1]的9个情况更新就可以了

if(j!=1) f[l][r][0][1]=(f[l][r][0][1]+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!=1) f[l][r][1][0]=(f[l][r][1][0]+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;

3.若l和r不是一对括号，那么往内分成两部分搜索，设l和p为一对括号

那么往内搜索（l,p）和(p+1,r)

因为当前l和r不是同一对括号，所以不需要像2中考虑同一对只能上一次色这个规则

只要考虑相邻颜色不同就可以了

f[l][r][i][j]=(f[l][r][i][j]+(f[l][p][i][k]*f[p+1][r][q][j])%MOD)%MOD;

在状态转移都结束之后，有一个问题就是3中的p如何计算，这里可以先预处理一下...

直接用栈的模拟处理出第i个括号相对的括号所在的位置就可以了...

用栈模拟括号应该是经典题了吧...碰到左括号就进栈，右括号就出栈

【代码】

#include <stdio.h>#include <string.h>#include <algorithm>#define MOD 1000000007using namespace std;int go[1000];int zhan[1000];long long f[750][750][3][3];void find(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(go[l]==r)    {        find(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(j!=2) f[l][r][0][2]=(f[l][r][0][2]+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(i!=2) f[l][r][2][0]=(f[l][r][2][0]+f[l+1][r-1][i][j])%MOD;            }    }    els    {        int p=go[l];        find(l,p);        find(p+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)))                        f[l][r][i][j]=(f[l][r][i][j]+(f[l][p][i][k]*f[p+1][r][q][j])%MOD)%MOD;    }}int main(){char s[1000];    scanf("%s",s);    int len=strlen(s);    int top=0;    for(int i=0;s[i]!='\0';i++)        if(s[i]=='(') zhan[top++]=i;        else        {       top--;        go[i]=zhan[top];            go[zhan[top]]=i;        }    memset(f,0,sizeof(f));    find(0,len-1);    long long 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;}