## solution

DP。頂点$1$からの距離別にやる。組合せ等を適当に前処理して$O(N^3)$。

## implementation

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define ALL(x) begin(x), end(x)
using namespace std;
template <typename X, typename T> auto vectors(X x, T a) { return vector<T>(x, a); }
template <typename X, typename Y, typename Z, typename... Zs> auto vectors(X x, Y y, Z z, Zs... zs) { auto cont = vectors(y, z, zs...); return vector<decltype(cont)>(x, cont); }

template <int32_t MOD>
struct mint {
int64_t data;  // faster than int32_t a little
mint() = default;  // data is not initialized
mint(int64_t value) : data(value) {}  // assume value is in proper range
inline mint<MOD> operator + (mint<MOD> other) const { int64_t c = this->data + other.data; return mint<MOD>(c >= MOD ? c - MOD : c); }
inline mint<MOD> operator - (mint<MOD> other) const { int64_t c = this->data - other.data; return mint<MOD>(c <    0 ? c + MOD : c); }
inline mint<MOD> operator * (mint<MOD> other) const { int64_t c = this->data * int64_t(other.data) % MOD; return mint<MOD>(c < 0 ? c + MOD : c); }
inline mint<MOD> & operator += (mint<MOD> other) { this->data += other.data; if (this->data >= MOD) this->data -= MOD; return *this; }
inline mint<MOD> & operator -= (mint<MOD> other) { this->data -= other.data; if (this->data <    0) this->data += MOD; return *this; }
inline mint<MOD> & operator *= (mint<MOD> other) { this->data = this->data * int64_t(other.data) % MOD; if (this->data < 0) this->data += MOD; return *this; }
inline mint<MOD> operator - () const { return mint<MOD>(this->data ? MOD - this->data : 0); }
mint<MOD> pow(uint64_t k) const {
mint<MOD> x = *this;
mint<MOD> y = 1;
for (uint64_t i = 1; i and (i <= k); i <<= 1) {
if (k & i) y *= x;
x *= x;
}
return y;
}
mint<MOD> inv() const {
return pow(MOD - 2);
}
};

template <int32_t MOD>
mint<MOD> fact(int n) {
static vector<mint<MOD> > memo(1, 1);
while (n >= memo.size()) {
memo.push_back(memo.back() * mint<MOD>(memo.size()));
}
return memo[n];
}
template <int32_t PRIME>
mint<PRIME> inv_fact(int n) {
static vector<mint<PRIME> > memo(1, 1);
while (n >= memo.size()) {
memo.push_back(memo.back() * mint<PRIME>(memo.size()).inv());
}
return memo[n];
}

template <int32_t MOD>
mint<MOD> choose(int n, int r) {
assert (0 <= r and r <= n);
return fact<MOD>(n) * inv_fact<MOD>(n - r) * inv_fact<MOD>(r);
}
int choose_two(int n) {
return n * (n - 1) / 2;
}

constexpr int MOD = 1e9 + 7;

int solve(int n, int d) {
const mint<MOD> two = 2;
mint<MOD> result = 0;
auto dp = vectors(n, n + 1, n + 1, mint<MOD>());
dp[0][1][1] = 1;
REP (i, n - 1) {
REP (j, (i + 1 >= d) ? n + 1 : n) {
dp[i + 1][j][0] = dp[i][j][0];
REP3 (k, 1, j) {
REP3 (l, 1, j - k + 1) {
auto new_node = choose<MOD>(n - (j - k) - (i + 1 <= d), k - (i + 1 == d));
auto new_edge_prv = (two.pow(l) - 1).pow(k);
auto new_edge_cur = two.pow(choose_two(k));
dp[i + 1][j][k] += dp[i][j - k][l] * new_node * new_edge_prv * new_edge_cur;
}
}
}
if (i + 1 >= d) {
REP (j, n + 1) {
auto new_graph = two.pow(choose_two(n - j));
result += accumulate(ALL(dp[i + 1][j]), mint<MOD>()) * new_graph;
}
}
}
return result.data;
}

int main() {
int n, d; cin >> n >> d;
cout << solve(n, d) << endl;
return 0;
}