#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll; typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename A,size_t N,typename T>void Fill(A(&array)[N],const T &val){fill((T*)array, (T*)(array+N), val);}
const int inf = 0x3fffffff; const ll INF = 0x3fffffffffffffff;
//template end

ll mod=1e9+7;
struct Mint {
    ll val;
    Mint inv() const{
        ll tmp,a=val,b=mod,x=1,y=0;
        while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
        return Mint(x);
    }
public:
    Mint():val(0){}
    Mint(ll x):val(x>=0?x%mod:x%mod+mod){}
    int mtoi(){return this->val; }
    Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
    Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
    Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
    Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;}
    Mint& operator/=(const Mint& x){return *this *= x.inv();}
    bool operator==(const Mint& x) const{return val == x.val;}
    bool operator!=(const Mint& x) const{return val != x.val;}
    bool operator<(const Mint& x) const{return val < x.val;}
    bool operator<=(const Mint& x) const{return val <= x.val;}
    bool operator>(const Mint& x) const{return val > x.val;}
    bool operator>=(const Mint& x) const{return val >= x.val;}
    Mint operator+(const Mint& x) const{return Mint(*this)+=x;}
    Mint operator-(const Mint& x) const{return Mint(*this)-=x;}
    Mint operator*(const Mint& x) const{return Mint(*this)*=x;}
    Mint operator/(const Mint& x) const{return Mint(*this)/=x;}
};
struct factorial {
    vector<Mint> Fact, Finv;
public:
    factorial(int maxx){
        Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*Mint(i+1);
        Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*Mint(i);
    }
    Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
    Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
    Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};

typedef tuple<vector<Mint>,vector<Mint>,Mint> func;
typedef tuple<Mint,Mint,Mint> T;
func to_f(T t){return func{{get<0>(t)},{get<1>(t)},get<2>(t)};}
void v_add(vector<Mint> &a,vector<Mint> b){
    a.resize(max(a.size(),b.size()));
    rep(i,0,b.size())a[i]+=b[i];
}
void v_dec(vector<Mint> &a,vector<Mint> b){
    a.resize(max(a.size(),b.size()));
    rep(i,0,b.size())a[i]-=b[i];
}
void t_add(T &a,T b){get<0>(a)+=get<0>(b),get<1>(a)+=get<1>(b),get<2>(a)+=get<2>(b);}
void t_dec(T &a,T b){get<0>(a)-=get<0>(b),get<1>(a)-=get<1>(b),get<2>(a)-=get<2>(b);}
void f_add(func &a,func b){v_add(get<0>(a),get<0>(b)),v_add(get<1>(a),get<1>(b)),get<2>(a)+=get<2>(b);}
func f_dec(func a,func b){v_dec(get<0>(a),get<0>(b)),v_dec(get<1>(a),get<1>(b)),get<2>(a)-=get<2>(b); return a;}
void integral(func &f){
    vector<Mint> p=get<0>(f),q=get<1>(f);
    if(p.back()!=Mint(0))p.resize(p.size()+1); 
    if(q.back()!=Mint(0))q.resize(q.size()+1);
    rrep(i,p.size()-1,0)p[i]=p[i-1]/Mint(i); p[0]=0;
    rrep(i,q.size()-1,0)q[i]=q[i-1]/Mint(i); q[0]=0;
    get<0>(f)=p; get<1>(f)=q; get<2>(f)*=-1;
}
func int_x(func f){integral(f); return func(f_dec(f,func{{get<2>(f)},{0},0}));}
T int_1(func f){
    integral(f); T res;
    vector<Mint> p=get<0>(f),q=get<1>(f);
    rep(i,0,p.size())get<0>(res)+=p[i];
    rep(i,0,q.size())get<1>(res)+=q[i];
    get<0>(res)-=get<2>(f); get<1>(res)+=get<2>(f); return res;
}
Mint pre[1010][2];
void calc(){
    pre[0][0]=1; pre[0][1]=-1;
    rep(i,0,1005){
        pre[i+1][0]=pre[i][0]*Mint(i+1);
        pre[i+1][1]=pre[i][1]*Mint(i+1)-1;
    }
}
T int_e(func f){
    T res; vector<Mint> p=get<0>(f),q=get<1>(f);
    rep(i,0,p.size())get<0>(res)+=p[i]*pre[i][0],get<1>(res)+=p[i]*pre[i][1];
    rep(i,0,q.size())get<1>(res)+=q[i]*pre[i][0],get<2>(res)+=q[i]*pre[i][1];
    get<0>(res)+=get<2>(f)/2; get<2>(res)-=get<2>(f)/2; return res;
}

func dp[1010][2][3];

int main(){
    int n; scanf("%d",&n);
    string s; cin>>s; calc(); Fill(dp,func{{0},{0},0});
    if(s[0]=='X')dp[0][0][0]=func{{0},{0},1};
    else dp[0][0][1]=func{{0},{0},1},dp[0][1][2]=func{{1},{0},-1};
    rep(i,1,n){
        if(s[i]=='X'){
            f_add(dp[i][0][0],int_x(dp[i-1][1][0]));
            f_add(dp[i][0][0],int_x(dp[i-1][1][2]));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][0][1])),int_x(dp[i-1][0][1])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][0][2])),int_x(dp[i-1][0][2])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][1][1])),int_x(dp[i-1][1][1])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][1][2])),int_x(dp[i-1][1][2])));
        }
        else{
            f_add(dp[i][0][1],int_x(dp[i-1][1][0]));
            f_add(dp[i][0][1],int_x(dp[i-1][1][2]));
            f_add(dp[i][1][2],int_x(dp[i-1][0][0]));
            f_add(dp[i][1][2],int_x(dp[i-1][0][2]));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][0][0])),int_x(dp[i-1][0][0])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][0][2])),int_x(dp[i-1][0][2])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][1][0])),int_x(dp[i-1][1][0])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][1][2])),int_x(dp[i-1][1][2])));
        }
    } T ans;
    rep(i,0,2){
        t_add(ans,int_e(dp[n-1][i][0]));
        t_add(ans,int_1(dp[n-1][i][1]));
        t_dec(ans,int_e(dp[n-1][i][1]));
        t_add(ans,int_1(dp[n-1][i][2]));
    }
    printf("%d %d %d\n",get<0>(ans).mtoi(),get<1>(ans).mtoi(),get<2>(ans).mtoi());
    return 0;
}
### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll; typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename A,size_t N,typename T>void Fill(A(&array)[N],const T &val){fill((T*)array, (T*)(array+N), val);}
const int inf = 0x3fffffff; const ll INF = 0x3fffffffffffffff;
//template end

ll mod=1e9+7;
struct Mint {
    ll val;
    Mint inv() const{
        ll tmp,a=val,b=mod,x=1,y=0;
        while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
        return Mint(x);
    }
public:
    Mint():val(0){}
    Mint(ll x):val(x>=0?x%mod:x%mod+mod){}
    int mtoi(){return this->val; }
    Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
    Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
    Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
    Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;}
    Mint& operator/=(const Mint& x){return *this *= x.inv();}
    bool operator==(const Mint& x) const{return val == x.val;}
    bool operator!=(const Mint& x) const{return val != x.val;}
    bool operator<(const Mint& x) const{return val < x.val;}
    bool operator<=(const Mint& x) const{return val <= x.val;}
    bool operator>(const Mint& x) const{return val > x.val;}
    bool operator>=(const Mint& x) const{return val >= x.val;}
    Mint operator+(const Mint& x) const{return Mint(*this)+=x;}
    Mint operator-(const Mint& x) const{return Mint(*this)-=x;}
    Mint operator*(const Mint& x) const{return Mint(*this)*=x;}
    Mint operator/(const Mint& x) const{return Mint(*this)/=x;}
};
struct factorial {
    vector<Mint> Fact, Finv;
public:
    factorial(int maxx){
        Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*Mint(i+1);
        Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*Mint(i);
    }
    Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
    Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
    Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};

typedef tuple<vector<Mint>,vector<Mint>,Mint> func;
typedef tuple<Mint,Mint,Mint> T;
func to_f(T t){return func{{get<0>(t)},{get<1>(t)},get<2>(t)};}
void v_add(vector<Mint> &a,vector<Mint> b){
    a.resize(max(a.size(),b.size()));
    rep(i,0,b.size())a[i]+=b[i];
}
void v_dec(vector<Mint> &a,vector<Mint> b){
    a.resize(max(a.size(),b.size()));
    rep(i,0,b.size())a[i]-=b[i];
}
void t_add(T &a,T b){get<0>(a)+=get<0>(b),get<1>(a)+=get<1>(b),get<2>(a)+=get<2>(b);}
void t_dec(T &a,T b){get<0>(a)-=get<0>(b),get<1>(a)-=get<1>(b),get<2>(a)-=get<2>(b);}
void f_add(func &a,func b){v_add(get<0>(a),get<0>(b)),v_add(get<1>(a),get<1>(b)),get<2>(a)+=get<2>(b);}
func f_dec(func a,func b){v_dec(get<0>(a),get<0>(b)),v_dec(get<1>(a),get<1>(b)),get<2>(a)-=get<2>(b); return a;}
void integral(func &f){
    vector<Mint> p=get<0>(f),q=get<1>(f);
    if(p.back()!=Mint(0))p.resize(p.size()+1); 
    if(q.back()!=Mint(0))q.resize(q.size()+1);
    rrep(i,p.size()-1,0)p[i]=p[i-1]/Mint(i); p[0]=0;
    rrep(i,q.size()-1,0)q[i]=q[i-1]/Mint(i); q[0]=0;
    get<0>(f)=p; get<1>(f)=q; get<2>(f)*=-1;
}
func int_x(func f){integral(f); return func(f_dec(f,func{{get<2>(f)},{0},0}));}
T int_1(func f){
    integral(f); T res;
    vector<Mint> p=get<0>(f),q=get<1>(f);
    rep(i,0,p.size())get<0>(res)+=p[i];
    rep(i,0,q.size())get<1>(res)+=q[i];
    get<0>(res)-=get<2>(f); get<1>(res)+=get<2>(f); return res;
}
Mint pre[1010][2];
void calc(){
    pre[0][0]=1; pre[0][1]=-1;
    rep(i,0,1005){
        pre[i+1][0]=pre[i][0]*Mint(i+1);
        pre[i+1][1]=pre[i][1]*Mint(i+1)-1;
    }
}
T int_e(func f){
    T res; vector<Mint> p=get<0>(f),q=get<1>(f);
    rep(i,0,p.size())get<0>(res)+=p[i]*pre[i][0],get<1>(res)+=p[i]*pre[i][1];
    rep(i,0,q.size())get<1>(res)+=q[i]*pre[i][0],get<2>(res)+=q[i]*pre[i][1];
    get<0>(res)+=get<2>(f)/2; get<2>(res)-=get<2>(f)/2; return res;
}

func dp[1010][2][3];

int main(){
    int n; scanf("%d",&n);
    string s; cin>>s; calc(); Fill(dp,func{{0},{0},0});
    if(s[0]=='X')dp[0][0][0]=func{{0},{0},1};
    else dp[0][0][1]=func{{0},{0},1},dp[0][1][2]=func{{1},{0},-1};
    rep(i,1,n){
        if(s[i]=='X'){
            f_add(dp[i][0][0],int_x(dp[i-1][1][0]));
            f_add(dp[i][0][0],int_x(dp[i-1][1][2]));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][0][1])),int_x(dp[i-1][0][1])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][0][2])),int_x(dp[i-1][0][2])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][1][1])),int_x(dp[i-1][1][1])));
            f_add(dp[i][0][0],f_dec(to_f(int_1(dp[i-1][1][2])),int_x(dp[i-1][1][2])));
        }
        else{
            f_add(dp[i][0][1],int_x(dp[i-1][1][0]));
            f_add(dp[i][0][1],int_x(dp[i-1][1][2]));
            f_add(dp[i][1][2],int_x(dp[i-1][0][0]));
            f_add(dp[i][1][2],int_x(dp[i-1][0][2]));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][0][0])),int_x(dp[i-1][0][0])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][0][2])),int_x(dp[i-1][0][2])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][1][0])),int_x(dp[i-1][1][0])));
            f_add(dp[i][0][1],f_dec(to_f(int_1(dp[i-1][1][2])),int_x(dp[i-1][1][2])));
        }
    } T ans;
    rep(i,0,2){
        t_add(ans,int_e(dp[n-1][i][0]));
        t_add(ans,int_1(dp[n-1][i][1]));
        t_dec(ans,int_e(dp[n-1][i][1]));
        t_add(ans,int_1(dp[n-1][i][2]));
    }
    printf("%d %d %d\n",get<0>(ans).mtoi(),get<1>(ans).mtoi(),get<2>(ans).mtoi());
    return 0;
}
```