// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long

#define ull unsigned ll

#define uint unsigned

#define pii pair<int,int>

#define pll pair<ll,ll>

#define PB push_back

#define fi first

#define se second

#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)

#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)

#define CLR(a,v) memset(a,v,sizeof(a));

#define CPY(a,b) memcpy(a,b,sizeof(a));

using namespace std;

const int mo=1000000007;

const int N=750;

int C[N][N],n,k,ans;

int used[N],q1[N],q2[N];

int suf[N],a[N];

char s[N];

void init(){

	For(i,0,N-1) C[i][0]=1;

	For(i,1,N-1) For(j,1,i)

		C[i][j]=(C[i-1][j]+C[i-1][j-1])%mo;

}

void solve(int S){

	if (!S) return ++ans,void(0);

	memset(used,0,sizeof(used));

	int rest=S-a[1],h=0,t=*q2=0;

	For(i,1,k)

		if (s[i]=='r') q1[++t]=i;

		else if (s[i]=='b'&&h!=t&&*q2!=rest)

			used[q1[++h]]=used[i]=1,q2[++*q2]=i;

	//printf("%d %d %d\n",rest,*q2,ans);

	//For(i,1,n) printf("%d ",a[i]); puts("");

	if (*q2!=rest) return;

	For(i,1,a[1])

		if (h==t) return;

		else used[q1[++h]]=1;

	//printf("%d %d %d\n",rest,*q2,ans);

	//For(i,1,max(n,k)) printf("%d ",a[i]); puts("");

	Rep(i,k,1) suf[i]=suf[i+1]+(!used[i]);

	For(i,1,S-a[1]) q2[i]=suf[q2[i]];

	int posi=S-a[1],sum=0,par=0;

	For(i,2,n) For(j,1,a[i]){

		sum+=i-2;

		if (sum>q2[posi]) return;

		--posi; 

	}

	int v=1,s=a[1];

	sum=a[1]+S-1;

	par=S+1+a[1];

	For(i,2,max(2,n)){

		v=1ll*v*C[s+a[i]][s]%mo;

		sum+=a[i]*(2*i-3);

		par+=a[i]*(2*i-1);

		s+=a[i]; 

	}

	if (sum>n) return;

	//printf("%d %d %d\n",sum,par,v);

	v=1ll*v*C[n-sum+par-1][par-1]%mo;

	ans=(ans+v)%mo;

}

/*



*/

void dfs(int x,int xx,int y,int S){

	if (y>x||2*max(y-1,1)>xx) return solve(S);

	for (a[y]=0;x>=0&&xx>=0;++a[y],++S)

		dfs(x,xx,y+1,S),x-=y,xx-=2*max(y-1,1);

	a[y]=0;

}

int main(){

	init();

	scanf("%d%d",&n,&k);

	scanf("%s",s+1);

	dfs(k,n+1,1,0);

	printf("%d\n",ans);

}

/*

40 4

rrbb

*/

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define ull unsigned ll

#define uint unsigned

#define pii pair<int,int>

#define pll pair<ll,ll>

#define PB push_back

#define fi first

#define se second

#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)

#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)

#define CLR(a,v) memset(a,v,sizeof(a));

#define CPY(a,b) memcpy(a,b,sizeof(a));

using namespace std;

const int mo=1000000007;

const int N=75;

int C[N*5][N*5];

int n,k,ans,used[N];

int q1[N],q2[N],suf[N];

int f[N][N];

char s[N];

void init(){

	For(i,0,N*5-1) C[i][0]=1;

	For(i,1,N*5-1) For(j,1,i)

		C[i][j]=(C[i-1][j]+C[i-1][j-1])%mo;

}

void solve(int S1,int S2){

	if (!S1&&!S2) return ++ans,void(0);

	if ((S1+S2)*2>n+1||S1+2*S2>k) return; 

	memset(used,0,sizeof(used));

	int h=0,t=*q2=0;

	For(i,1,k)

		if (s[i]=='r') q1[++t]=i;

		else if (s[i]=='b'&&h!=t&&*q2!=S2)

			used[q1[++h]]=used[i]=1,q2[++*q2]=i;

	if (*q2!=S2) return;

	For(i,1,S1)

		if (h==t) return;

		else used[q1[++h]]=1;

	Rep(i,k,1) suf[i]=suf[i+1]+(!used[i]);

	For(i,1,S2) q2[i]=suf[q2[i]];

	memset(f,0,sizeof(f));

	f[0][0]=1;

	int upp=max(2,n)+1;

	For(i,2,upp) Rep(j,S2-1,0) For(k,2*j,upp-i*(S2-j)){

		int psum=k-2*j;

		For(l,1,min(S2-j,(upp-k)/i)){

			if ((psum+=i-2)>q2[S2-j-l+1]) break;

			f[j+l][k+i*l]=(f[j+l][k+i*l]+1ll*C[j+l][l]*f[j][k])%mo;

		}

	}

	For(k,0,upp){

		int sum=2*S1+2*k-2*S2-1;

		int par=2*S1+2*k+1;

		if (sum&&sum<=n&&f[S2][k])

			ans=(ans+1ll*f[S2][k]*C[n-sum+par-1][par-1]%mo*C[S2+S1][S2])%mo;

	}

}

int main(){

	init();

	scanf("%d%d",&n,&k);

	scanf("%s",s+1);

	For(S1,0,n) For(S2,0,n) solve(S1,S2);

	printf("%d\n",ans);

}

// Fastest known solution for this problem: