There is a very long bench. The bench is divided into M sections, where M is a very large integer.

Initially, the bench is vacant. Then, M people come to the bench one by one, and perform the following action:

* We call a section comfortable if the section is currently unoccupied and is not adjacent to any occupied sections. If there is no comfortable section, the person leaves the bench. Otherwise, the person chooses one of comfortable sections uniformly at random, and sits there. (The choices are independent from each other).



After all M people perform actions, Snuke chooses an interval of N consecutive sections uniformly at random (from M-N+1 possible intervals), and takes a photo. His photo can be described by a string of length N consisting of `X` and `-`: the i-th character of the string is `X` if the i-th section from the left in the interval is occupied, and `-` otherwise. Note that the photo is directed. For example, `-X--X` and `X--X-` are different photos.

What is the probability that the photo matches a given string s? This probability depends on M. You need to compute the limit of this probability when M goes infinity.

Here, we can prove that the limit can be uniquely written in the following format using three rational numbers p, q, r and e = 2.718 \ldots (the base of natural logarithm):

p + \frac{q}{e} + \frac{r}{e^2}

Your task is to compute these three rational numbers, and print them modulo 10^9 + 7, as described in Notes section.

Constraints

* 1 \leq N \leq 1000
* |s| = N
* s consists of `X` and `-`.

Input

Input is given from Standard Input in the following format:


N
s


Output

Print three rational numbers p, q, r, separated by spaces.

Examples

Input

1
X


Output

500000004 0 500000003


Input

3
---


Output

0 0 0


Input

5
X--X-


Output

0 0 1


Input

5
X-X-X


Output

500000004 0 833333337


Input

20
-X--X--X-X--X--X-X-X


Output

0 0 183703705


Input

100
X-X-X-X-X-X-X-X-X-X--X-X-X-X-X-X-X-X-X-X-X-X-X-X-X--X--X-X-X-X--X--X-X-X--X-X-X--X-X--X--X-X--X-X-X-


Output

0 0 435664291
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// gp_hash_table<int, int> mapka;

using namespace std;
#define PB push_back
#define MP make_pair
#define LL long long
#define int LL
#define FOR(i,a,b) for(int i = (a); i <= (b); i++)
#define RE(i,n) FOR(i,1,n)
#define REP(i,n) FOR(i,0,(int)(n)-1)
#define R(i,n) REP(i,n)
#define VI vector<int>
#define PII pair<int,int>
#define LD long double
#define FI first
#define SE second
#define st FI
#define nd SE
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())

template<class C> void mini(C &a4, C b4) { a4 = min(a4, b4); }
template<class C> void maxi(C &a4, C b4) { a4 = max(a4, b4); }

template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
  while(*sdbg!=',')cerr<<*sdbg++;
  cerr<<'='<<h<<','; _dbg(sdbg+1, a...);
}

template<class T> ostream &operator<<(ostream& os, vector<T> V) {
  os << "["; for (auto vv : V) os << vv << ","; return os << "]";
}
template<class L, class R> ostream &operator<<(ostream &os, pair<L,R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}

#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) (__VA_ARGS__)
#define cerr if(0)cout
#endif


const int P = 1e9 + 7;
void add(int &a,int b){
  a += b;
  if(a >= P)
    a -= P;
}
void sub(int &a,int b){
  a -= b;
  if(a < 0)
    a += P;
}

int mul(int a,int b){
  return a * b % P;
}
struct war{
  int a,b,c;
  void operator +=(war x){
    add(a, x.a);
    add(b, x.b);
    add(c, x.c);
  }
  void operator -=(war x){
    sub(a, x.a);
    sub(b, x.b);
    sub(c, x.c);
  }
  
  void operator *=(war x){
    c = mul(c, x.a);
    add(c,mul(a, x.c));
    add(c,mul(b, x.b));
    
    
    b = mul(b, x.a);
    add(b,mul(a, x.b));
    
    a = mul(a, x.a);
  }
};

const int MAX = 1001;
int dp[MAX][3];

war licz1(int x){
  war res = {0,1,0};
  R(i,x - 1){
    res -= {0,dp[i + 1][2],0};
  }
  debug(x, res.b * 3 % P);
  return res;
}

war licz2(int x){
  debug(x);
  if(x == 1){
    return {(P + 1) / 2, 0, (P - 1) / 2};
  }
  war res = {1,0,0};
  res -= licz2(x-1);
  R(i,2 * x - 2){
    war pom = licz1((i + 1) / 2);
    pom *= {0,1,0};
    res -= pom;
  }
  return res;
}

int pot(int a,int w){
  int r = 1;
  while(w){
    if(w & 1){
      r = r * a % P;
    }
    w/=2;
    a = a * a % P;
  }
  return r;
}

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  cerr << fixed << setprecision(6);
  
  dp[0][1] = dp[0][0] = 1;
  dp[0][2] = 0;
  R(i,MAX)if(i)R(j,3){
    R(k,i){
      add(dp[i][j], mul(pot(2 * i - 1 + j, P - 2), mul(dp[k][j / 2], dp[i - k - 1][(j + 1) / 2])));
    }
  }
  debug(dp[0][2]);
  debug(dp[1][2] * 3 % P);
  debug(dp[2][2] * 5 % P);
  debug(dp[3][2] * 7 % P);
  
  int n;
  string z;
  cin >> n >> z;
  if(n == 1 && z[0] == '-'){
    cout << (P + 1) / 2 << " 0 " << (P + 1) / 2 << "\n";
    return 0;
  }
  R(i,n-1)if(z[i] == 'X' && z[i+1] == 'X'){
    cout << "0 0 0\n";
    return 0;
  }
  int ile = 0;
  war res = {1,0,0};
  vector<int> x;
  R(i,n){
    if(z[i] == '-'){
      if(i && z[i-1] == '-'){
        x.PB(ile);
        ile = 0;
      }
    }else{
      ile ++;
    }
  }
  debug(ile, x);
  if(x.empty()){
    res = licz2(ile);
  }else{
    res *= licz1(ile);
    res *= licz1(x[0]);
    R(i,SZ(x))if(i){
      res *= {dp[x[i]][2],0,0};
    }
  }
  cout << res.a << " " << res.b << " " << res.c << "\n";
}
