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

// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) long long(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;

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; }

const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;

ll n; 
ll a[2005];

ll g(ll a, ll b, ll c){
  if(b==0 && c==0) return 0;
  if(a*b*c==1) return 0;
  return 1;
}

// 1段目の1,2個目がv0,v1の時の通り数
ll f(ll v0, ll v1){
  vector<vvll> dp(n+1, vvll(2, vll(2, 0))); // 
  dp[2][v0][v1]=1;
  
  // 1段目の0~nの状態に対して2段目の状態と辻褄が合うかを検査しており、この段階ではa[0],a[n-1]と辻褄が合うかはわからない
  for(ll i=2; i<n; i++) rep(j,2) rep(k,2) rep(l,2){
    if(g(j,k,l)==a[i-1]) (dp[i+1][k][l]+=dp[i][j][k])%=MOD;
  }
  
  ll ans=0;
  rep(i,2) rep(j,2){
    if(g(i,j,v0)==a[n-1] && g(j,v0,v1)==a[0]) (ans+=dp[n][i][j])%=MOD;
  }
  return ans;
}



signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  
  cin>>n;
  rep(i,n) cin>>a[i];
  
  ll ans=0;
  rep(v0, 2) rep(v1, 2) (ans+=f(v0, v1))%=MOD;
  
  cout<<ans<<endl;
  
}