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:
/*
Author: QAQAutomaton
Lang: C++
Code: F.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
template<class T>bool sort2(T &a,T &b){return a>b?swap(a,b),1:0;}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		fill((unsigned char*)&inf,(unsigned char*)&inf+sizeof(inf),0x3f);
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
const int p=1000000007;
struct Z{
	uint x;
	Z(){}
	Z(uint a){
		x=a;
	}
};
inline uint modp(const uint x){
	return x<p?x:x-p;
}
inline Z operator+(const Z x1, const Z x2) { return modp(x1.x+x2.x);}
inline Z operator-(const Z x1, const Z x2) { return modp(x1.x+p-x2.x);}
inline Z operator-(const Z x) {return x.x?p-x.x:0;}
inline Z operator*(const Z x1, const Z x2) { return static_cast<ull>(x1.x)*x2.x%p;}
void exgcd(int a,int b,int &x,int &y){
	if(!b){x=1;y=0;return;}
	exgcd(b,a%b,y,x);
	y-=(a/b)*x;
}
inline Z Inv(const Z a){
	int x,y;
	exgcd(p,a.x,x,y);
	return y<0?y+=p:y;
}
inline Z operator/(const Z x1, const Z x2) { return x1*Inv(x2);}

inline Z &operator++(Z &x1){x1.x==p-1?x1.x=0:++x1.x;return x1;} 
inline Z &operator--(Z &x1){x1.x?--x1.x:x1.x=p-1;return x1;}
inline Z &operator+=(Z &x1, const Z x2) { return x1 = x1 + x2; }
inline Z &operator-=(Z &x1, const Z x2) { return x1 = x1 - x2; }
inline Z &operator*=(Z &x1, const Z x2) { return x1 = x1 * x2; }
inline Z &operator/=(Z &x1, const Z x2) { return x1 = x1 / x2; }
inline Z fpm(Z a,int b){Z c(1);for(;b;b>>=1,a*=a)if(b&1)c*=a;return c;}
Z ans;
int cnt=0;
int n,k;
int m;
int a[75];
int b[75];
char s[75];
Z c[1005][1005];
void check(){
//	for(int i=1;i<=m;++i)write(a[i],i==m?'\n':' ');
	int c1=1,c2=1;
	for(int i=1;i<=m;++i){
		b[i]=a[i]+1;
		c1+=(a[i]+1)<<1;
		c2+=(a[i]?2:0);
	}
	if(m)++c2;
	int at1=m,at2=m,at3=m;
	for(int i=1;i<=k;++i){
		if(s[i]=='r'){
			if(at1){--b[at1];--at1;}
			else{
				while(at3>at2 && !b[at3])--at3;
				if(at3>at2){--b[at3];}
			}
		}
		else{
			if(at2>at1 && b[at2]){--b[at2];--at2;}
			else{
				while(at3>at2 && !b[at3])--at3;
				if(at3>at2){--b[at3];}
			}
		}
	}
	for(int i=1;i<=m;++i)if(b[i])return;
	Z w=c[n+c2-1][c1-1];
	for(int i=1;i<=m;){
		int j=i;
		while(j<=m && a[j]==a[i])++j;
		w*=c[j-1][j-i];
		i=j;
	}
//	write(n,' ',c1,' ',c2,' ',c[n+c2-1][c1-1].x,'|',w.x,",ok!\n");
	ans+=w;
/*	for(int i=1;i<=m;++i)write(a[i],i==m?'\n':' ');
	write('|',c1,' ',c2,'|',k+c2-1,',',c1-1,' ',c[k+c2-1][c1-1].x,'\n');*/
}
void dfs(int x,int y){
	check();
	for(;y<=x;++y){
		a[++m]=y;
		dfs(x-y,y);
		--m;
	}
}
signed main(){
#ifdef QAQAutoMaton 
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	read(n,k);
	for(int i=0;i<=1000;++i){
		c[i][0]=c[i][i]=1;
		for(int j=1;j<i;++j)c[i][j]=c[i-1][j]+c[i-1][j-1];
	}
	read(s+1);
	int s=(n+1)>>1;
	while(s>=0){
		dfs(s,1);
		a[++m]=0;
		--s;
	}
	write(ans.x,'\n');
	return 0;
}