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:
#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
using namespace std;

void mod_inverse(int n, vector<long long>& inv, int mod)
{
    inv.assign(n+1, -1);
    inv[1] = 1;
    for(int i=2; i<=n; ++i)
        inv[i] = inv[mod % i] * (mod - mod / i) % mod;
}

class FactorialCalculation
{
private:
    const int mod;
    vector<long long> factorial;
    vector<long long> invFactorial;
public:
    FactorialCalculation(int n, int mod) : mod(mod)
    {
        factorial.resize(n+1, 1);
        invFactorial.resize(n+1, 1);
        vector<long long> inv;
        mod_inverse(n, inv, mod);
        for(int i=1; i<=n; ++i){
            factorial[i] = factorial[i-1] * i % mod;
            invFactorial[i] = invFactorial[i-1] * inv[i] % mod;
        }
    }
    long long getFactorial(int n){
        return factorial[n];
    }
    long long getInvFactorial(int n){
        return invFactorial[n];
    }
    long long getPermutation(int n, int r){
        if(n < r)
            return 0;
        return factorial[n] * invFactorial[n-r] % mod;
    }
    long long getCombination(int n, int r){
        if(n < r)
            return 0;
        return getPermutation(n, r) * invFactorial[r] % mod;
    }
    long long getHomogeneous(int n, int r){
        return getCombination(n+r-1, r);
    }
};

const long long MOD = 1000000007;
FactorialCalculation fc(200, MOD);

int ballNum, opeNum; // ボールの個数、操作列の長さ
string opeStr; // 操作列

long long calc(const vector<int>& v)
{
    int n = v.size();
    int cnt = 1;
    int sum = 1;
    long long ans = 1;
    for(int i=0; i<n; ++i){
        if(i < n - 1 && v[i] == v[i+1]){
            ++ cnt;
        }
        else{
            ans *= fc.getHomogeneous(sum, cnt);
            ans %= MOD;
            sum += cnt;
            cnt = 1;
        }
    }
    return ans;
}

long long solve(const vector<int>& v)
{
    int size = v.size();
    vector<int> index(size+1, -1);
    vector<bool> check(opeNum, false);
    for(int i=size-1; i>=0; --i){
        index[i] = index[i+1] + 1;
        while(index[i] < opeNum && opeStr[index[i]] != 'r')
            ++ index[i];
        if(index[i] == opeNum)
            return -1;
        check[index[i]] = true;
    }
    for(int i=size-1; i>=0; --i){
        if(v[i] > 1){
            while(index[i] < opeNum && (check[index[i]] || opeStr[index[i]] != 'b'))
                ++ index[i];
            if(index[i] == opeNum)
                return -1;
            check[index[i]] = true;
        }
    }
    for(int i=size-1; i>=0; --i){
        for(int j=3; j<=v[i]; ++j){
            while(index[i] < opeNum && check[index[i]])
                ++ index[i];
            if(index[i] == opeNum)
                return -1;
            check[index[i]] = true;
        }
    }

    int cnt = v.size() + 1;
    int rest = ballNum - (v.size() - 1);
    for(int i=0; i<size; ++i){
        cnt += v[i] * 2 - 1;
        rest -= max(1, v[i] * 2 - 3);
    }
    if(rest < 0)
        return -1;

    long long ans = fc.getHomogeneous(cnt, rest);
    ans *= calc(v);
    ans %= MOD;
    return ans;
}

long long dfs(int a, vector<int> v)
{
    long long ans = 0;
    for(;;){
        v.push_back(a);
        long long tmp = solve(v);
        if(tmp == -1)
            break;
        ans += tmp;
        ans %= MOD;

        v.pop_back();
        ans += dfs(a+1, v);
        ans %= MOD;
        v.push_back(a);
    }
    return ans;
}

int main()
{
    cin >> ballNum >> opeNum >> opeStr;

    long long ans = dfs(1, vector<int>()) + 1;
    ans %= MOD;
    cout << ans << endl;
}