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
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define FILL(a, n, x) memset((a), (x), sizeof ((a)[0]) * (n))
#define COPY(a, n, b) memcpy((b), (a), sizeof ((a)[0]) * (n))
using namespace std;

typedef long long li;

const int mod = 1e9 + 7;
inline int Add(int x) { return x >= mod ? x - mod : x; }
inline void Add(int &x, int y) { x += y; if (x >= mod) x -= mod; }
inline int Sub(int x) { return x < 0 ? x + mod : x; }
inline void Sub(int &x, int y) { x -= y; if (x < 0) x += mod; }
inline int Mul(int x, int y) { return (li)x * y % mod; }

// ----------------------------------------

const int maxn = 2e5;

int n, m, lim;
char s[maxn + 1];
int dp[maxn + 1];
int sum[maxn + 2];

void GetLim(void) {
  int cnt = 0;
  for (int i = 0; i < m; ++i) {
    if (s[i] == 'R') ++cnt; else break;
  }
  if (cnt & 1) lim = cnt; else lim = cnt + 1;
  cnt = 0;
  for (int i = 0; i < m; ++i) {
    if (s[i] == 'R') ++cnt; else {
      if (cnt & 1) lim = min(lim, cnt);
      cnt = 0;
    }
  }
}

void SolveSame(void) {
  static int dp[maxn + 1][2][2];
  dp[0][0][0] = dp[0][1][1] = 1;
  for (int i = 1; i < n; ++i) {
    for (int x = 0; x <= 1; ++x) {
      for (int y = 0; y <= 1; ++y) {
        for (int z = 0; z <= 1; ++z) {
          if (y != 1 || z != 1) {
            Add(dp[i][x][z], dp[i - 1][x][y]);
          }
        }
      }
    }
  }
  int ans = Add(Add(dp[n - 1][0][0] + dp[n - 1][0][1]) + dp[n - 1][1][0]);
  printf("%d\n", ans);
}

int main(void) {
  scanf("%d%d%s", &n, &m, s);
  if (s[0] == 'B') {
    for (int i = 0; i < m; ++i) {
      s[i] ^= 'R' ^ 'B';
    }
  }
  if (count(s, s + m, 'R') == m) {
    // the same
    SolveSame();
    return 0;
  }

  GetLim();
  lim = lim + 1 >> 1;
  if (n & 1) {
    puts("0");
    return 0;
  }
  n /= 2;

  for (int i = 1; i <= lim; ++i) {
    Add(dp[i], 2 * i);
  }
  for (int i = 1; i <= n; ++i) {
    // printf("this is %d\n", i);
    // calc dp[i]
    int L = max(1, i - lim), R = i;
    Add(dp[i], Sub(sum[R] - sum[L]));
    sum[i + 1] = Add(sum[i] + dp[i]);
  }
  printf("%d\n", dp[n]);
  return 0;
}
