## solution

DP。$k$のときの長さの総和の期待値(答え) $f(k)$と、 $k$のときの赤い頂点の深さの平均の期待値$g(k)$を求める。 根で止まるひとつを除いた$k - 1$個がどう左右に振り分けられるかを総当たり。$O(k)$。

## implementation

#include <bits/stdc++.h>
#include <tuple>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
typedef long long ll;
using namespace std;
class RndSubTree { public: int count(int k); };

ll powmod(ll x, ll y, ll m) {
assert (0 <= x and x < m);
assert (0 <= y);
ll z = 1;
for (ll i = 1; i <= y; i <<= 1) {
if (y & i) z = z * x % m;
x = x * x % m;
}
return z;
}
ll modinv(ll x, ll p) {
assert (x % p != 0);
return powmod(x, p - 2, p);
}
template <int32_t MOD>
int32_t fact(int n) {
static vector<int32_t> memo(1, 1);
while (n >= memo.size()) {
memo.push_back(memo.back() *(int64_t) memo.size() % MOD);
}
return memo[n];
}
template <int32_t PRIME>
int32_t inv_fact(int n) {
static vector<int32_t> memo(1, 1);
while (n >= memo.size()) {
memo.push_back(memo.back() *(int64_t) modinv(memo.size(), PRIME) % PRIME);
}
return memo[n];
}
template <int MOD>
int choose(int n, int r) {
assert (0 <= r and r <= n);
return fact<MOD>(n) *(ll) inv_fact<MOD>(n - r) % MOD *(ll) inv_fact<MOD>(r) % MOD;
}

constexpr int MOD = 1e9 + 7;
pair<ll, ll> addpq(pair<ll, ll> r1, pair<ll, ll> r2) {
ll p1, q1; tie(p1, q1) = r1;
ll p2, q2; tie(p2, q2) = r2;
ll p = (p1 * q2 % MOD + p2 * q1 % MOD) % MOD;
ll q = q1 * q2 % MOD;
return make_pair(p, q);
}
pair<ll, ll> mulpq(pair<ll, ll> r1, pair<ll, ll> r2) {
ll p1, q1; tie(p1, q1) = r1;
ll p2, q2; tie(p2, q2) = r2;
ll p = p1 * p2 % MOD;
ll q = q1 * q2 % MOD;
return make_pair(p, q);
}
pair<pair<ll, ll>, pair<ll, ll> > recur(int k) {
if (k == 0) {
return make_pair(make_pair(0, 1), make_pair(0, 1));
} else if (k == 1) {
return make_pair(make_pair(0, 1), make_pair(1, 1));
} else {
static map<int, pair<pair<ll, ll>, pair<ll, ll> > > memo;
if (memo.count(k)) return memo[k];
pair<ll, ll> acc = { 0, 1 };
pair<ll, ll> acc_up = { 0, 1 };
ll den = powmod(2, k - 1, MOD);
REP (k1, k) {
int k2 = (k - 1) - k1;
ll num = choose<MOD>(k1 + k2, k1);
pair<ll, ll> a1, up1; tie(a1, up1) = recur(k1);
pair<ll, ll> a2, up2; tie(a2, up2) = recur(k2);
pair<ll, ll> a = { 0, 1 };
a = addpq(a, mulpq(up1, make_pair(k1, 1)));
a = addpq(a, mulpq(up2, make_pair(k2, 1)));
a = mulpq(a, make_pair(num, den));
pair<ll, ll> up = { 0, 1 };
up = addpq(up, mulpq(up1, make_pair(k1, 1)));
up = addpq(up, mulpq(up2, make_pair(k2, 1)));
up = mulpq(up, make_pair(1, k));