Problem: There are N white balls arranged in a row, numbered 1,2,..,N from left to right. AtCoDeer the deer is thinking of painting some of these balls red and blue, while leaving some of them white.

You are given a string s of length K. AtCoDeer performs the following operation for each i from 1 through K in order:

* The i-th operation: Choose a contiguous segment of balls (possibly empty), and paint these balls red if the i-th character in s is `r`; paint them blue if the character is `b`.



Here, if a ball which is already painted is again painted, the color of the ball will be overwritten. However, due to the properties of dyes, it is not possible to paint a white, unpainted ball directly in blue. That is, when the i-th character in s is `b`, the chosen segment must not contain a white ball.

After all the operations, how many different sequences of colors of the balls are possible? Since the count can be large, find it modulo 10^9+7.

Constraints

* 1 ≤ N ≤ 70
* 1 ≤ K ≤ 70
* |s| = K
* s consists of `r` and `b`.
* N and K are integers.

Input

Input is given from Standard Input in the following format:


N K
s


Output

Print the number of the different possible sequences of colors of the balls after all the operations, modulo 10^9+7.

Examples

Input

2 2
rb


Output

9


Input

5 2
br


Output

16


Input

7 4
rbrb


Output

1569


Input

70 70
bbrbrrbbrrbbbbrbbrbrrbbrrbbrbrrbrbrbbbbrbbrbrrbbrrbbbbrbbrbrrbbrrbbbbr


Output

841634130

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<iostream>
#include<map>
#include<string>
#include<bitset>

#define MOD 1000000007

using namespace std;

typedef long long ll;

inline void add(int &x,int y) {
  ((x+=y)>=MOD)?x-=MOD:0;
} 

ll C[105][105];

void pre(int n) {
  for(int i=0;i<=n;i++) C[i][0]=1;
  for(int i=1;i<=n;i++)
    for(int j=1;j<=i;j++) C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
}

int f[85][85][85],up[85];
int sum[85];

void dp(int n,int m,int c) {
  memset(f,0,sizeof(f));
  int t=m+2*c-1;
  f[0][0][0]=1;
  for(int i=0;i<m;i++)
    for(int j=0;j<=up[i]&&2*j-m+t<=n;j++) {
      int s=f[i][j][0];
      for(int k=1;k<=up[m]&&j+(m-i)*k<=up[m];k++) {
      	for(int l=1;i+l<=m&&j+l*k<=up[i+l];l++)
      	  add(f[i+l][j+l*k][k],s*C[i+l][l]%MOD);
      	add(s,f[i][j][k]);
	  }
    }
  for(int i=0;i<=up[m]&&2*i-m+t<=n;i++) {
  	int s=0;
  	for(int j=0;j<=i;j++) add(s,f[m][i][j]);
  	if (!s) continue;
  	int v=2*i-m+t;
  	for(int j=0;j<=2*m+2&&v+j<=n;j++)
  	  add(sum[v+j],s*C[m+c][c]%MOD*C[2*m+2][j]%MOD);
  }
}

char str[85];

int q[85],lpos[85];
bool vis[85];

int main() {
  int n,k;
  scanf("%d%d%s",&n,&k,str+1);
  pre(100);
  for(int i=0;2*i<=k;i++) {
  	memset(vis,0,sizeof(vis));
  	int cnt=0;
  	int lx=1,rx=0;
  	for(int j=1;j<=k&&cnt<i;j++)
  	  if (str[j]=='r') q[++rx]=j;
  	  else if (lx<=rx) {
  	  	  vis[q[lx++]]=vis[j]=1;
  	  	  lpos[++cnt]=j;
		}
	if (cnt<i) continue;
	int s=0;
	for(int j=1;j<=k;j++)
	  if (!vis[j]&&str[j]=='r') s++;
	int r=1;
	lpos[i+1]=k+1;up[i+1]=0;
	for(int j=0;j<=s;j++) {
		if (j) {
			while (vis[r]||str[r]!='r') r++;
			vis[r]=1;
		}
		for(int t1=i;t1>0;t1--) {
			up[t1]=up[t1+1];
			for(int t2=lpos[t1+1]-1;t2>lpos[t1];t2--)
			  if (!vis[t2]) up[t1]++;
			up[t1]++;
		}
		reverse(up+1,up+i+1);
		dp(n,i,j);
	}
  }
  int ans=0;
  for(int i=1;i<=n;i++) add(ans,sum[i]*C[n-1][i-1]%MOD);
  printf("%d\n",ans);
  return 0;
}