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:xht37
#include <bits/stdc++.h>

#define ui unsigned int
#define ll long long
#define ul unsigned ll
#define ld long double

#define pi pair <int, int>
#define fi first
#define se second
#define mp make_pair

#define ls (p << 1)
#define rs (ls | 1)
#define md ((t[p].l + t[p].r) >> 1)

#define vi vector <int>
#define pb push_back
#define pq priority_queue

#define dbg(x) cerr << #x" = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)

#define fl(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)

using namespace std;

namespace io {
	const int SI = 1 << 21 | 1;
	char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
	int f, t;
	#define gc() (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin), IS == IT ? EOF : *IS++) : *IS++)
	inline void flush() {
		fwrite(OB, 1, OS - OB, stdout), OS = OB;
	}
	inline void pc(char x) {
		*OS++ = x;
		if (OS == OT) flush();
	}

	template <class I>
	inline void rd(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15), c = gc());
		x *= f;
	}
	template <class I>
	inline void rd(I &x, I &y) {
		rd(x), rd(y);
	}
	template <class I>
	inline void rd(I &x, I &y, I &z) {
		rd(x), rd(y), rd(z);
	}
	template <class I>
	inline void rda(I *a, int n) {
		for (int i = 1; i <= n; i++) rd(a[i]);
	}
	inline void rdc(char &c) {
		for (c = gc(); c < 33 || c > 126; c = gc());
	}
	inline void rds(char *s, int &n) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (n = 0; c >= 33 && c <= 126; s[++n] = c, c = gc());
		s[n+1] = '\0';
	}
	inline void rds(string &s) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (s.clear(); c >= 33 && c <= 126; s.pb(c), c = gc());
	}

	template <class I>
	inline void print(I x, char k = '\n') {
		if (!x) pc('0');
		if (x < 0) pc('-'), x = -x;
		while (x) ch[++t] = x % 10 + '0', x /= 10;
		while (t) pc(ch[t--]);
		pc(k);
	}
	template <class I>
	inline void print(I x, I y) {
		print(x, ' '), print(y);
	}
	template <class I>
	inline void print(I x, I y, I z) {
		print(x, ' '), print(y, ' '), print(z);
	}
	template <class I>
	inline void printa(I *a, int n) {
		for (int i = 1; i <= n; i++) print(a[i], " \n"[i==n]);
	}
	inline void printc(char c) {
		pc(c);
	}
	inline void prints(char *s, int n) {
		for (int i = 1; i <= n; i++) pc(s[i]);
		pc('\n');
	}
	inline void prints(string s) {
		int n = s.length();
		while (t < n) pc(s[t++]);
		pc('\n'), t = 0;
	}
	struct Flush {
		~Flush() {
			flush();
		}
	} flusher;
}
using io::rd;
using io::rda;
using io::rdc;
using io::rds;
using io::print;
using io::printa;
using io::printc;
using io::prints;

const int P = 1e9 + 7;

struct modint {
    int x;
    inline modint(int x = 0) : x(x) {}
    inline modint &operator = (int o) { return x = o, *this; }
    inline modint &operator += (modint o) { return (x += o.x) >= P && (x -= P), *this; }
    inline modint &operator -= (modint o) { return (x -= o.x) < 0 && (x += P), *this; }
    inline modint &operator *= (modint o) { return x = 1ll * x * o.x % P, *this; }
    inline modint &operator ^= (ll b) {
        modint a = *this, c;
        if (!~b) b = P - 2;
        c.x = 1 % P;
        while (b) {
            if (b & 1) c *= a;
            a *= a, b >>= 1;
        }
        return x = c.x, *this;
    }
    inline modint &operator /= (modint o) { return *this *= o ^ -1; }
    inline modint &operator += (int o) { return (x += o) >= P && (x -= P), *this; }
    inline modint &operator -= (int o) { return (x -= o) < 0 && (x += P), *this; }
    inline modint &operator *= (int o) { return x = 1ll * x * o % P, *this; }
    inline modint &operator /= (int o) { return *this *= (modint)o ^ -1; }
    template <class I>
    inline friend modint operator + (modint a, I b) { return a += b; }
    template <class I>
    inline friend modint operator - (modint a, I b) { return a -= b; }
    template <class I>
    inline friend modint operator * (modint a, I b) { return a *= b; }
    inline friend modint operator ^ (modint a, ll b) { return a ^= b; }
    template <class I>
    inline friend modint operator / (modint a, I b) { return a /= b; }
    inline friend bool operator == (modint a, int b) { return a.x == b; }
    inline friend bool operator != (modint a, int b) { return a.x != b; }
    inline friend bool operator < (modint a, int b) { return a.x < b; }
    inline friend bool operator <= (modint a, int b) { return a.x <= b; }
    inline friend bool operator > (modint a, int b) { return a.x > b; }
    inline friend bool operator >= (modint a, int b) { return a.x >= b; }
    inline friend bool operator == (modint a, modint b) { return a.x == b.x; }
    inline friend bool operator != (modint a, modint b) { return a.x != b.x; }
    inline friend bool operator < (modint a, modint b) { return a.x < b.x; }
    inline friend bool operator <= (modint a, modint b) { return a.x <= b.x; }
    inline friend bool operator > (modint a, modint b) { return a.x > b.x; }
    inline friend bool operator >= (modint a, modint b) { return a.x >= b.x; }
    inline bool operator ! () { return !x; }
    inline modint operator - () { return x ? P - x : 0; }
};
inline void rd(modint &x) { rd(x.x); }
inline void print(modint x, char k = '\n') { print(x.x, k); }

const int NP = 1e6 + 7;
modint p[NP], v[NP], vp[NP];
inline void init(int n) {
  p[0] = v[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = p[i-1] * i;
  vp[n] = 1 / p[n];
  for (int i = n; i; i--) v[i] = vp[i] * p[i-1], vp[i-1] = vp[i] * i;
}
inline modint binom(int n, int m) {
  return n < m ? 0 : p[n] * vp[m] * vp[n-m];
}

int n, k;
string s;
modint ans;

inline bool pd(vi p) {
	vector<bool> v(k, 0);
	vi t(p.size(), -1);
	for (ui i = 0; i < p.size(); i++) {
		while (++t[i] < k && (v[t[i]] || s[t[i]] == 'b'));
		if (t[i] == k) return 0;
		v[t[i]] = 1;
	}
	while (p.size() && p.back() == 1) p.pop_back();
	for (ui i = 0; i < p.size(); i++) {
		while (++t[i] < k && (v[t[i]] || s[t[i]] == 'r'));
		if (t[i] == k) return 0;
		v[t[i]] = 1;
	}
	while (p.size() && p.back() == 2) p.pop_back();
	for (ui i = 0; i < p.size(); i++)
		for (int j = 2; j < p[i]; j++) {
			while (++t[i] < k && v[t[i]]);
			if (t[i] == k) return 0;
			v[t[i]] = 1;
		}
	return 1;
}

void dfs(int o, int l, int r, vi p) {
	if (!o) {
		if (!pd(p)) return;
		modint now = binom(n - l + r - 1, r - 1);
		while (p.size()) {
			int t = 0, x = p.back();
			while (p.size() && p.back() == x) ++t, p.pop_back();
			now *= binom(p.size() + t, t); 
		}
//		debug("l = %d, r = %d\n", l, r);
//		for (int x : p) debug("%d ", x);
//		debug("\n%d\n", now.x);
		return ans += now, void();
	}
	while (1) {
		dfs(o - 1, l, r, p);
		l += (o == 1 ? 1 : 2 * o - 3) + (l != 0);
		if (l > n) return;
		r += 2 * o;
		p.pb(o);
	}
}

int main() {
	rd(n, k), rds(s);
	init(1e6);
	vi p;
	dfs((n + 3) >> 1, 0, 1, p);
	print(ans);
	return 0;
}