You are at the top left cell (1, 1) of an n × m labyrinth. Your goal is to get to the bottom right cell (n, m). You can only move right or down, one cell per step. Moving right from a cell (x, y) takes you to the cell (x, y + 1), while moving down takes you to the cell (x + 1, y).

Some cells of the labyrinth contain rocks. When you move to a cell with rock, the rock is pushed to the next cell in the direction you're moving. If the next cell contains a rock, it gets pushed further, and so on.

The labyrinth is surrounded by impenetrable walls, thus any move that would put you or any rock outside of the labyrinth is illegal.

Count the number of different legal paths you can take from the start to the goal modulo 10^9 + 7. Two paths are considered different if there is at least one cell that is visited in one path, but not visited in the other.

Input

The first line contains two integers n, m — dimensions of the labyrinth (1 ≤ n, m ≤ 2000).

Next n lines describe the labyrinth. Each of these lines contains m characters. The j-th character of the i-th of these lines is equal to "R" if the cell (i, j) contains a rock, or "." if the cell (i, j) is empty.

It is guaranteed that the starting cell (1, 1) is empty.

Output

Print a single integer — the number of different legal paths from (1, 1) to (n, m) modulo 10^9 + 7.

Examples

Input


1 1
.


Output


1


Input


2 3
...
..R


Output


0


Input


4 4
...R
.RR.
.RR.
R...


Output


4

Note

In the first sample case we can't (and don't have to) move, hence the only path consists of a single cell (1, 1).

In the second sample case the goal is blocked and is unreachable.

Illustrations for the third sample case can be found here: <https://assets.codeforces.com/rounds/1225/index.html>
#include <bits/stdc++.h>
using namespace std;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &x : v) is >> x;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (i != v.size() - 1) cout << endl;
  };
  return os;
}
template <typename T>
void Out(T x) {
  cout << x << endl;
}
template <typename T1, typename T2>
void Ans(bool first, T1 y, T2 n) {
  if (first)
    Out(y);
  else
    Out(n);
}
using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;
const int mod = 1000000007;
template <long long MOD>
struct ModInt {
  using ll = long long;
  ll val;
  void setval(ll v) { val = v % MOD; };
  ModInt() : val(0) {}
  ModInt(ll v) { setval(v); };
  ModInt operator+(const ModInt &x) const { return ModInt(val + x.val); }
  ModInt operator-(const ModInt &x) const { return ModInt(val - x.val + MOD); }
  ModInt operator*(const ModInt &x) const { return ModInt(val * x.val); }
  ModInt operator/(const ModInt &x) const { return *this * x.inv(); }
  ModInt operator-() const { return ModInt(MOD - val); }
  ModInt operator+=(const ModInt &x) { return *this = *this + x; }
  ModInt operator-=(const ModInt &x) { return *this = *this - x; }
  ModInt operator*=(const ModInt &x) { return *this = *this * x; }
  ModInt operator/=(const ModInt &x) { return *this = *this / x; }
  friend ostream &operator<<(ostream &os, const ModInt &x) {
    os << x.val;
    return os;
  }
  friend istream &operator>>(istream &is, ModInt &x) {
    is >> x.val;
    x.val = (x.val % MOD + MOD) % MOD;
    return is;
  }
  ModInt pow(ll n) const {
    ModInt a = 1;
    if (n == 0) return a;
    int i0 = 64 - __builtin_clzll(n);
    for (int i = i0 - 1; i >= 0; i--) {
      a = a * a;
      if ((n >> i) & 1) a *= (*this);
    }
    return a;
  }
  ModInt inv() const { return this->pow(MOD - 2); }
};
using mint = ModInt<mod>;
mint pow(mint x, long long n) { return x.pow(n); }
using mvec = vector<mint>;
using mmat = vector<mvec>;
struct Combination {
  vector<mint> fact, invfact;
  Combination(int N) {
    fact = vector<mint>({mint(1)});
    invfact = vector<mint>({mint(1)});
    fact_initialize(N);
  }
  void fact_initialize(int N) {
    int i0 = fact.size();
    if (i0 >= N + 1) return;
    fact.resize(N + 1);
    invfact.resize(N + 1);
    for (int i = i0; i <= N; i++) fact[i] = fact[i - 1] * i;
    invfact[N] = (mint)1 / fact[N];
    for (int i = N - 1; i >= i0; i--) invfact[i] = invfact[i + 1] * (i + 1);
  }
  mint nCr(int n, int r) {
    if (n < 0 || r < 0 || r > n) return mint(0);
    if (fact.size() < n + 1) fact_initialize(n);
    return fact[n] * invfact[r] * invfact[n - r];
  }
  mint nPr(int n, int r) {
    if (n < 0 || r < 0 || r > n) return mint(0);
    if (fact.size() < n + 1) fact_initialize(n);
    return fact[n] * invfact[n - r];
  }
};
signed main() {
  int n, m;
  cin >> n >> m;
  vector<string> second(n);
  cin >> second;
  if (second[0][0] == 'R' || second[n - 1][m - 1] == 'R') {
    Out(0);
    return 0;
  }
  if (n == 1 && m == 1) {
    Out(1);
    return 0;
  }
  mat cnt1(n, vec(m + 1, 0)), cnt2(n + 1, vec(m, 0));
  mat ls1(n, vec(m + 1, 0)), ls2(m, vec(n + 1, 0));
  for (int i = (0); i < (n); i++) {
    for (int j = (m)-1; j >= (0); j--) {
      cnt1[i][j] = cnt1[i][j + 1];
      if (second[i][j] == 'R') {
        cnt1[i][j]++;
        ls1[i][cnt1[i][j]] = j;
      }
    }
  }
  for (int j = (0); j < (m); j++) {
    for (int i = (n)-1; i >= (0); i--) {
      cnt2[i][j] = cnt2[i + 1][j];
      if (second[i][j] == 'R') {
        cnt2[i][j]++;
        ls2[j][cnt2[i][j]] = i;
      }
    }
  }
  mmat dp1(n, mvec(m, 0)), dp2(n, mvec(m, 0));
  mmat S1(n, mvec(m + 1, 0)), S2(n + 1, mvec(m, 0));
  dp1[0][0] = 1;
  dp2[0][0] = 1;
  S1[0][1] = 1;
  S2[1][0] = 1;
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (m); j++)
      if (i > 0 || j > 0) {
        if (j > 0) {
          int M = m - 1 - j;
          int j0 = ls1[i][M + 1];
          dp2[i][j] = S1[i][j] - S1[i][j0];
        }
        if (i > 0) {
          int M = n - 1 - i;
          int i0 = ls2[j][M + 1];
          dp1[i][j] = S2[i][j] - S2[i0][j];
        }
        S1[i][j + 1] = S1[i][j] + dp1[i][j];
        S2[i + 1][j] = S2[i][j] + dp2[i][j];
      }
  mint ans = dp1[n - 1][m - 1] + dp2[n - 1][m - 1];
  Out(ans);
  return 0;
}
