cf# 149 D Coloring Brackets (interval DP)

D-Coloring BracketsTime limit:$ MS Memory Limit:262144KB 64bit IO Format:%i64d &%i64u SubmitStatusPracticecodeforces 149D

Description

Once Petya read a problem about a bracket sequence. He gave it much thought but didn ' t find a solution. Today you'll 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's possible to obtain a correct mathematical expression from it, Inserting numbers and operators between the brackets. For example, such sequences as " (()) () " and " ( ) is correct bracket sequences and such sequences as ") () " and " (() "is not.

In a correct bracket sequence all bracket corresponds to the matching bracket (an opening bracket corresponds to the MATC Hing 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 is allowed to color some brackets in the bracket sequence so as all three conditions is fulfilled:

• Each bracket is either isn't colored any color, or is colored red, or is colored blue.
• For any pair of matching brackets exactly one of the them is colored. In other words, for any bracket the following are true:either it or the matching bracket that corresponds to it is colored .
• No. Neighboring colored brackets has the same color.

Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. The ways of coloring is considered different if they differ in the color of least one bracket. As the result can be quite large, print it modulo 1000000007 (^9?+?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, meet the above given conditions modulo 1 000000007 (^9?+?7).

Sample Input

Input

(())

Output

12

Input

(()())

Output

40

Input

()

Output

4

Hint

Let ' s consider the first sample test. The bracket sequence from the sample can being colored, for example, as was shown on, and figures below.

The ways of coloring shown below are incorrect.

Reference Link: http://blog.csdn.net/sdjzping/article/details/19160013

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <queue> #include <stack> #include <vector> #include <set> #include <map > #define L (x) (x<<1) #define R (x) (x<<1|1) #define MID (x, y) ((x+y) >>1) #define EPS 1e-8typedef __ Int64 ll; #define FRE (i,a,b) for (i = A; I <b; i++) #define FREE (i,b,a) for (i = b; I >= a;i--) #define MEM (T, v) MEMS ET ((t), V, sizeof (t)) #define SSF (n) scanf ("%s", N) #define SF (n) scanf ("%d", &n) #define SFF (A, b) scanf (          "%d%d", &a, &b) #define SFFF (a,b,c) scanf ("%d%d%d", &a, &b, &c) #define PF Printf#define Bug PF ("hi\n") using namespace std; #define MOD 1000000007#define INF 0x3f3f3f3f#define n 705ll Dp[n][n][3][3];char c[n ];int to[n];int len;void inint () {int i,j,p=0;int t[n];for (i=0;i<len;i++) if (c[i]== ' (') t[p++]=i;else{to[i]=to[t[ p-1]];to[t[p-1]]=i;p--;}} void dfs (int le,int ri) {if (Le>=ri) rEturn;     if (Le+1==ri) {dp[le][ri][0][1]=1; dp[le][ri][1][0]=1; dp[le][ri][0][2]=1; dp[le][ri][2][0]=1; return;}     int i,j; if (To[le]==ri) {DFS (le+1,ri-1); for (i=0;i<3;i++) for (j=0;j<3;j++) {if (j!=1) dp[le][ri][0][1]= (dp[le][ri][0][   1]+DP[LE+1][RI-1][I][J])%mod;   if (i!=1) dp[le][ri][1][0]= (dp[le][ri][1][0]+dp[le+1][ri-1][i][j])%mod;   if (j!=2) dp[le][ri][0][2]= (dp[le][ri][0][2]+dp[le+1][ri-1][i][j])%mod;   if (i!=2) dp[le][ri][2][0]= (dp[le][ri][2][0]+dp[le+1][ri-1][i][j])%mod; } return; } int P=to[le]; DFS (LE,P); DFS (P+1,RI); int II,JJ; for (i=0;i<3;i++) for (j=0;j<3;j++) for (ii=0;ii<3;ii++) for (jj=0;jj<3;jj++) {if (j==1&&ii==1| | j==2&&ii==2) continue; Dp[le][ri][i][jj]= (dp[le][ri][i][jj]+ (DP[LE][P][I][J]*DP[P+1][RI][II][JJ])%mod)%mod;     }}int Main () {int i,j;    while (~SCANF ("%s", C)) {Len=strlen (c); Inint ();    DFS (0,LEN-1);    ll Ans=0; for (i=0;i<3;i++) for (j=0;j<3;j++) ans= (ans+dp[0][len-1][i][j])%mod; printf"%i64d\n", ans); } return 0;}

cf# 149 D Coloring Brackets (interval DP)