CF 149 D Coloring Brackets（区间dp）

D. 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 modulo1000000007 (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.

题意：给定一个匹配的括号序列，对于每一对括号，必须对其中一个括号染色，可以染红色或者蓝色，相邻括号要么同时无色要么为不同颜色。现在要求一共有多少种染色方案。

思路：给定的括号为已经匹配了的合法序列，dp[l][r][i][j]表示区间[l,r]两段的颜色分别为i,j的时候有多少种染色方案。0,1,2分别代表不染色，染红色和染蓝色。

对于l+1==r的情况，也就是最小匹配括号对，四种染色情况都为1

对于l和r匹配的情况，根据相邻括号颜色需要不同来分类讨论

最后l和r不匹配的情况，预处理出每一个左括号匹配的右括号的位置，如果遇到l和r不匹配的情况的话，那就将区间[l,r]分成两段递归。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
typedef long long ll;
const int maxn=705;
const int mod=1e9+7;
ll dp[maxn][maxn][4][4];
int len,top;
ll st[maxn],m[maxn];
char s[maxn];
void match()    
{

    for(int i=1;i<=len;i++)
    {
        if(s[i]=='(') st[++top]=i;
        else
        {
            int temp=st[top--];
            m[temp]=i;
            m[i]=temp;
        }
    }
}
void dfs(int l,int r)
{
    if(l>=r) return;
    if(l+1==r)
    {
        dp[l][r][1][0]=1;
        dp[l][r][0][1]=1;
        dp[l][r][2][0]=1;
        dp[l][r][0][2]=1;
        return;
    }
    if(m[l]==r)
    {
        dfs(l+1,r-1);
        for(int i=0;i<3;i++)
            for(int j=0;j<3;j++)
            {
                if(j!=1) dp[l][r][0][1]=(dp[l][r][0][1]+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;
                if(i!=1) dp[l][r][1][0]=(dp[l][r][1][0]+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;
            }
    }
    else
    {
        int temp=m[l];
        dfs(l,temp);
        dfs(temp+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 t=0;t<3;t++)
                    {
                        if(k==1&&t==1) continue;
                        if(k==2&&t==2) continue;
                        dp[l][r][i][j]=(dp[l][r][i][j]+dp[l][temp][i][k]*dp[temp+1][r][t][j]%mod)%mod;
                    }
    }
}
int main()
{
    scanf("%s",s+1);
    len=strlen(s+1);
    top=0;
    match();
    ll ans=0;
    dfs(1,len);
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
        {
            ans=(ans+dp[1][len][i][j])%mod;
        }
    printf("%lld\n",ans);

}