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<bits/stdc++.h>
using namespace std;
const int MAXN = 75;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int fac[MAXN], inv[MAXN], invn[MAXN];
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
int binom(int x, int y) {
	if (y > x) return 0;
	else return 1ll * fac[x] * inv[y] % P * inv[x - y] % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
char s[MAXN];
int n, m, p[MAXN], q[MAXN], r[MAXN], sum[MAXN];
unordered_map <int, int> dp[MAXN][MAXN][MAXN];
int split(int x, int y) {
	if (y > x) return 0;
	else return binom(x - 1, y - 1);
}
inline int Hash(int x, int y) {
	return x * MAXN + y;
}
int getdp(int lft, int pos, int opt, int last, int cnt) {
	if (pos == 0) return lft <= 1;
	if (last > opt + 1) return 0;
	if ((last + 1) * pos > lft) return 0;
	if (last * pos > opt + sum[pos]) return 0;
	if (dp[lft][pos][opt].count(Hash(last, cnt))) {
		return dp[lft][pos][opt][Hash(last, cnt)];
	}
	int &ans = dp[lft][pos][opt][Hash(last, cnt)];
	ans = getdp(lft, pos, opt, last + 1, 0);
	if (last == 0) {
		if (lft >= 2) {
			update(ans, 1ll * invn[cnt + 1] * 
			                  getdp(lft - 2, pos - 1, opt + r[pos] + (q[pos] != 0), last, cnt + 1) % P);
		}
	} else if (q[pos] != 0) {
		int base = 2 * last;
		if (lft >= base) {
			update(ans, 1ll * invn[cnt + 1] *
			                  getdp(lft - base, pos - 1, opt + r[pos] + 1 - last, last, cnt + 1) % P);
		}
		if (lft >= base + 1) {
			update(ans, 2ll * invn[cnt + 1] *
			                  getdp(lft - base - 1, pos - 1, opt + r[pos] + 1 - last, last, cnt + 1) % P);
		}
		if (lft >= base + 2) {
			update(ans, 1ll * invn[cnt + 1] *
			                  getdp(lft - base - 2, pos - 1, opt + r[pos] + 1 - last, last, cnt + 1) % P);
		}
	}
	return ans;
}
int func(int cnt) {
	static bool vis[MAXN]; int lft = 0;
	memset(vis, false, sizeof(vis));
	for (int i = 1; i <= m; i++)
		if (s[i] == 'r' && lft != cnt) {
			p[++lft] = i, q[lft] = 0;
			vis[i] = true;
			for (int j = i; j <= m; j++)
				if (!vis[j] && s[j] == 'b') {
					vis[j] = true;
					q[lft] = j;
					break;
				}
		}
	if (lft != cnt) return 0;
	for (int i = 1; i <= m; i++) {
		r[i] = 0;
		for (int j = q[i - 1] + 1; j <= q[i] - 1; j++)
			r[i] += !vis[j];
		sum[i] = sum[i - 1] + r[i] + vis[p[i]] + vis[q[i]];
	}
	int cmt = 0;
	for (int i = m; i >= 1; i--)
		if (vis[i]) break;
		else cmt++;
	for (int i = 0; i < MAXN; i++)
	for (int j = 0; j < MAXN; j++)
	for (int k = 0; k < MAXN; k++)
		dp[i][j][k].clear();
	int ans = 0;
	for (int i = 1; i <= n; i++)
		update(ans, 1ll * (getdp(i, cnt, cmt, 0, 0) + getdp(i + 1, cnt, cmt, 0, 0)) * split(n, i) % P);
	return 1ll * ans * fac[cnt] % P;
}
int main() {
	read(n), read(m);
	scanf("\n%s", s + 1);
	int ans = 0;
	fac[0] = inv[0] = 1;
	for (int i = 1; i <= max(n, m); i++) {
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = power(fac[i], P - 2);
		invn[i] = power(i, P - 2);
	}
	for (int i = 0; i <= m; i++)
		update(ans, func(i));
	writeln(ans);
	return 0;
}