#pragma GCC optimize("O3") #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; using P = pair; using T = tuple; template inline T chmax(T &a, const T b) {return a = (a < b) ? b : a;} template inline T chmin(T &a, const T b) {return a = (a > b) ? b : a;} constexpr int MOD = 1e9 + 7; constexpr int inf = 1e9; constexpr long long INF = 1e18; #define all(a) (a).begin(), (a).end() int dx[] = {1, 0, -1, 0}; int dy[] = {0, 1, 0, -1}; ll modpow(ll a, ll b){ if(b == 0) return 1; else if(b % 2 == 0){ ll d = modpow(a, b / 2) % MOD; return (d * d) % MOD; } else{ return (a * modpow(a, b - 1)) % MOD; } } int main(){ cin.tie(0); ios::sync_with_stdio(false); int n; cin>>n; ll p; cin>>p; vector a(n + 1, 0); a[2] = 1; for(int i=3; i<=n; i++){ a[i] = (p * a[i - 1]) % MOD + a[i - 2]; a[i] %= MOD; } ll asum = accumulate(all(a), 0LL) % MOD; ll ans = 0; for(int i=2; i<=n; i++){ ans += a[i] * a[i] % MOD; ans %= MOD; } for(int i=2; i<=n; i++){ ans += a[i] * asum % MOD; ans %= MOD; } cout << ans * modpow(2, MOD - 2) % MOD << endl; return 0; }