Hakone Ekiden is one of the Japanese New Year's traditions. In Hakone Ekiden, 10 runners from each team aim for the goal while connecting the sashes at each relay station. In the TV broadcast, the ranking change from the previous relay station is displayed along with the passing order of each team at the relay station. So, look at it and tell us how many possible passage orders for each team at the previous relay station. The number of possible transit orders can be very large, so answer by the remainder divided by 1,000,000,007.



Input

The input is given in the form:

> n
> c1
> c2
> ...
> cn
>

The number n (1 ≤ n ≤ 200) representing the number of teams is on the first line, and the ranking changes from the previous relay station in order from the first place on the following n lines. If it is'`U`', the ranking is up, if it is'`-`', the ranking is not changed) is written.

Output

Please output in one line by dividing the number of passages that could have been the previous relay station by 1,000,000,007.

Examples

Input

3
-
U
D


Output

1


Input

5
U
U
-
D
D


Output

5


Input

8
U
D
D
D
D
D
D
D


Output

1


Input

10
U
D
U
D
U
D
U
D
U
D


Output

608


Input

2
D
U


Output

0
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
const int inf = 1e9 + 7;
const ll INF = 1e18;
const int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }


void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);

    int T;
    // cin >> T;
    T = 1;

    while (T--) {
        solve();
    }
}

// https://drken1215.hatenablog.com/entry/2019/10/05/173700
void solve() {
  int n;
  cin >> n;
  vector<char> a(n);
  cin >> a;

  // 上から i 段目, すでに j 個マッチングしてる
  int dp[300][300];
  fill_n(*dp, 300 * 300, 0);
  dp[0][0] = 1;

  rep(i, n) rep(j, n + 1) {
    if (a[i] == '-') {
      add(dp[i + 1][j + 1], dp[i][j]);
    } else if (a[i] == 'D') {
      // 頂点 D からは, 上に上がってる (相方 (前回の中継所) からが下がってる)
      // ペアは上がるか下がるか

      // i-j : 上にあるのが i 個
      add(dp[i + 1][j + 2], 1LL * dp[i][j] * (i - j) * (i - j) % mod);
      add(dp[i + 1][j + 1], 1LL * dp[i][j] * (i - j) % mod);
    } else {
      add(dp[i + 1][j], dp[i][j]);
      add(dp[i + 1][j + 1], 1LL * dp[i][j] * (i - j) % mod);
    }
  }
  cout << dp[n][n] << endl;
  // rep(i, n + 1) { rep(j, n + 1) cout << dp[i][j] << " "; cout << endl; }
}

