#include #include #include #include //#include #include #include #include #include #include //#include #include #include #include //#include #include #include //#include #include #include #include #include const int dx[] = {1, 0, -1, 0}; const int dy[] = {0, 1, 0, -1}; using namespace std; typedef long long ll; typedef vector vi; typedef vector vll; typedef pair pii; typedef long long number; typedef vector vec; typedef vector matrix; const ll MOD = 1000; // O( n ) matrix identity(int n) { matrix A(n, vec(n)); for (int i = 0; i < n; ++i) A[i][i] = 1; return A; } // O( n^3 ) matrix mul(const matrix &A, const matrix &B) { matrix C(A.size(), vec(B[0].size())); for (int i = 0; i < (int)C.size(); ++i) for (int j = 0; j < (int)C[i].size(); ++j) for (int k = 0; k < (int)A[i].size(); ++k) { C[i][j] += A[i][k] * B[k][j]; C[i][j] %= MOD; } return C; } // O( n^3 log e ) matrix pow(const matrix &A, ll e) { if (e == 0) return identity(A.size()); if (e == 1) return A; if (e % 2 == 0) { matrix tmp = pow(A, e/2); return mul(tmp, tmp); } else { matrix tmp = pow(A, e-1); return mul(A, tmp); } } int main() { cin.tie(0); ios::sync_with_stdio(false); int n; cin >> n; matrix mat(2, vec(2)); mat[0][0] = 1; mat[0][1] = 3; mat[1][0] = 1; mat[1][1] = 1; mat = pow(mat, n); ll ans = mat[0][0]*2; if (n%2 == 0) ans = (ans-1+MOD); cout << ans%MOD << endl; return 0; }