Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:

* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.



Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.

You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.

Note that the rotations of the same coloring are also distinguished.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.

Input

Input is given from Standard Input in the following format:


N M
S


Output

Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.

Examples

Input

4 7
RBRRBRR


Output

2


Input

3 3
BBB


Output

4


Input

12 10
RRRRBRRRRB


Output

78
def solve(n, m, s):
    f = s[0]
    MOD = 10 ** 9 + 7
    INF = 10 ** 6

    p = None
    seq = 0
    min_seq = INF
    for c in s:
        if c == p:
            seq += 1
        else:
            if p == f and (min_seq == INF or seq % 2 == 1):
                min_seq = min(min_seq, seq)
            seq = 1
            p = c

    if min_seq == INF:
        a, b = 1, 0
        for _ in range(n - 2):
            a, b = (a + b) % MOD, a
        return (3 * a + b) % MOD

    if n % 2 == 1:
        return 0

    n2 = n // 2
    ms = min(n2, min_seq // 2 + 1)

    dp = [0] * (n2 + 1)
    dp[0] = 1
    acc, reject = 1, 0
    for i in range(1, n2 + 1):
        dp[i] = (acc - reject) % MOD
        acc = (acc + dp[i]) % MOD
        if i >= ms:
            reject = (reject + dp[i - ms]) % MOD

    ans = 0
    for d in range(1, ms + 1):
        ans = (ans + dp[n2 - d] * d * 2) % MOD

    return ans


n, m = map(int, input().split())
s = input()

print(solve(n, m, s))
