Problem: Let's assume that set S consists of m distinct intervals [l1, r1], [l2, r2], ..., [lm, rm] (1 ≤ li ≤ ri ≤ n; li, ri are integers).

Let's assume that f(S) is the maximum number of intervals that you can choose from the set S, such that every two of them do not intersect. We assume that two intervals, [l1, r1] and [l2, r2], intersect if there is an integer x, which meets two inequalities: l1 ≤ x ≤ r1 and l2 ≤ x ≤ r2.

Sereja wonders, how many sets S are there, such that f(S) = k? Count this number modulo 1000000007 (109 + 7).

Input

The first line contains integers n, k (1 ≤ n ≤ 500; 0 ≤ k ≤ 500).

Output

In a single line, print the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

3 1


Output

23


Input

3 2


Output

32


Input

2 0


Output

1


Input

2 2


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool upmin(T &x, T y) {
  return y <= x ? x = y, 1 : 0;
}
template <typename T>
inline bool upmax(T &x, T y) {
  return x <= y ? x = y, 1 : 0;
}
const long double eps = 1e-9;
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long loo = 1ll << 60;
const int MAXN = 300005;
const int MAXM = 4005;
const int MX = 200000;
const int G = 3;
const int mods = 1e9 + 7;
const int Gi = (mods + 1) / G;
const int SZ = 62;
const int inv2 = (mods + 1) >> 1;
const int INF = 0x3f3f3f3f;
namespace FastIO {
constexpr int SIZE = (1 << 21) + 1;
int num = 0, f;
char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf,
                                                *oT = obuf + SIZE - 1, c;
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
}
inline void putc(char c) {
  *oS++ = c;
  if (oS == oT) flush();
}
inline void getc(char &c) {
  for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++);
       !isalpha(c) && c != EOF;
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    ;
}
inline void reads(char *st) {
  char c;
  int n = 0;
  getc(st[++n]);
  for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++);
       isalpha(c);
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    st[++n] = c;
  st[++n] = '\0';
}
template <class I>
inline void read(I &x) {
  for (f = 1, c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                              (iS == iT ? EOF : *iS++))
                            : *iS++);
       c < '0' || c > '9';
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9';
       c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    x = (x << 3) + (x << 1) + (c & 15);
  x *= f;
}
template <class I>
inline void print(I x) {
  if (x < 0) putc('-'), x = -x;
  if (!x) putc('0');
  while (x) que[++num] = x % 10 + 48, x /= 10;
  while (num) putc(que[num--]);
}
inline void putstr(string st) {
  for (int i = 0; i < (int)st.size(); ++i) putc(st[i]);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_Flusher_;
}  // namespace FastIO
using FastIO ::getc;
using FastIO ::print;
using FastIO ::putc;
using FastIO ::putstr;
using FastIO ::read;
using FastIO ::reads;
long long pw[MAXN], f[505][505];
signed main() {
  int n, m;
  read(n), read(m), pw[0] = 1;
  for (int i = 1; i <= n * n; ++i) pw[i] = (pw[i - 1] << 1) % mods;
  for (int i = 0; i <= n; ++i) f[i][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= min(m, i); ++j)
      for (int k = j - 1; k <= i; ++k)
        f[i][j] = (f[i][j] + f[k][j - 1] * (pw[i - k] - 1) % mods *
                                 pw[k * (i - k)] % mods) %
                  mods;
  print(f[n][m]);
  return 0;
}