#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define _GLIBCXX_DEBUG
using namespace std;
using std::cout;
using std::cin;
using std::endl;
using ll=long long;
using ld=long double;
ll ILL=1167167167167167167;
const int INF=100000000;
const ll mod=1e9+7;
#define rep(i,a) for (ll i=0;i<a;i++)
template<class T> using _pq = priority_queue<T, vector<T>, greater<T>>;
template<class T> ll LB(vector<T> &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> ll UB(vector<T> &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}else return 0;}
template<class T> bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}else return 0;}
template<class T> void So(vector<T> &v) {sort(v.begin(),v.end());}
template<class T> void Sore(vector<T> &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});}
void yneos(bool a){if(a) cout<<"Yes\n"; else cout<<"No\n";}
// return val=p(N)
// a=p[0].first^p[0].second * ... *p[N-1].first^p[N-1].second
// for all i: p[i].first is prime number
// O(sqrt(val))
std::vector<std::pair<long long,long long>> Prime_factorization(long long val){
	assert(val>=1);
	if(val==1){
		return {};
	}
	int ind=0;
	std::vector<std::pair<long long,long long>> ans;
	for(long long i=2;i*i<=val;i++){
		if(val%i!=0) continue;
		ans.push_back({i,0});
		while(val%i==0){
			ans[ind].second++;
			val/=i;
		}
		ind++;
	}
	if(val!=1) ans.push_back({val,1});
	return ans;
}

ll jyo(ll x,ll y,ll z){
  ll H=y; //ここから
       ll a=1,b=x,c=1;
       while(H>0){
         a*=2;
         if(H%a!=0){
           H-=a/2;
           c*=b;
           c%=z;
         }
        b*=b;
         b%=z;
      } //ここまで
 return c;
}


 
void solve();

//  rainy ~ 雨に打たれて ~
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int t=1;
	//cin>>t;
	rep(i,t) solve();
}

void solve(){
	ll N;
	string K;
	cin>>N>>K;
	vector<ll> p={0,1,1,2,1,2,2,3};
	ll ans=0;
	ll A=jyo(4,N-1,mod),B=jyo(2,N-1,mod);
	if(N==1){
		int a=0;
		rep(i,8) if(K[i]=='o') a++;
		cout<<a<<"\n";
	}
	rep(i,8){
		if(K[i]=='o'){
			ans+=(A*((4*(B+1)+(B-1)*(p[i]-2)*(p[i]-2))%mod))%mod;
		}
	}
	cout<<(ans%mod+mod)%mod<<"\n";
}