Problem: Let's define the transformation P of a sequence of integers a1, a2, ..., an as b1, b2, ..., bn, where bi = a1 | a2 | ... | ai for all i = 1, 2, ..., n, where | is the bitwise OR operation.

Vasya consequently applies the transformation P to all sequences of length n consisting of integers from 1 to 2k - 1 inclusive. He wants to know how many of these sequences have such property that their transformation is a strictly increasing sequence. Help him to calculate this number modulo 109 + 7.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 1 ≤ k ≤ 30 000).

Output

Print a single integer — the answer to the problem modulo 109 + 7.

Examples

Input

1 2


Output

3


Input

2 3


Output

30


Input

3 3


Output

48

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 31000;
long long n, k, len, rev[1 << 18], fac[N], inv[N];
inline void add(long long &a, long long b, long long mod) {
  a = ((a + b >= mod) ? a + b - mod : a + b);
}
inline void del(long long &a, long long b, long long mod) {
  a = ((a - b < 0) ? a - b + mod : a - b);
}
inline void mul(long long &a, long long b, long long mod) { a = a * b % mod; }
inline long long m_pow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) mul(ans, a, mod);
    b >>= 1;
    mul(a, a, mod);
  }
  return ans;
}
struct mint {
  long long a, b, c;
  inline long long get() {
    long long b1 = (b - a + 1004535809) *
                   m_pow(998244353, 1004535809 - 2, 1004535809) % 1004535809,
              a4, b4;
    a4 = (a + b1 * 998244353) % (1ll * 998244353 * 1004535809);
    b4 = (c - a4 % 469762049 + 469762049) *
         m_pow(1ll * 998244353 * 1004535809 % 469762049, 469762049 - 2,
               469762049) %
         469762049;
    b4 %= 1000000007;
    mul(b4, 998244353, 1000000007);
    mul(b4, 1004535809, 1000000007);
    add(b4, a4 % 1000000007, 1000000007);
    return b4;
  }
  mint &operator=(long long x) {
    a = b = c = x;
    return *this;
  }
};
mint operator+(mint a, mint b) {
  add(a.a, b.a, 998244353);
  add(a.b, b.b, 1004535809);
  add(a.c, b.c, 469762049);
  return a;
}
mint operator-(mint a, mint b) {
  del(a.a, b.a, 998244353);
  del(a.b, b.b, 1004535809);
  del(a.c, b.c, 469762049);
  return a;
}
mint operator*(mint a, mint b) {
  mul(a.a, b.a, 998244353);
  mul(a.b, b.b, 1004535809);
  mul(a.c, b.c, 469762049);
  return a;
}
mint operator/(mint a, long long b) {
  mul(a.a, m_pow(b, 998244353 - 2, 998244353), 998244353);
  mul(a.b, m_pow(b, 1004535809 - 2, 1004535809), 1004535809);
  mul(a.c, m_pow(b, 469762049 - 2, 469762049), 469762049);
  return a;
}
vector<mint> w[19];
inline long long read() {
  long long f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void prework() {
  long long d = 18;
  for (long long i = 1; i <= d; i++) {
    w[i].resize(1 << i);
    w[i][0] = (mint){1, 1, 1};
    w[i][1] = (mint){m_pow(3, (998244353 - 1) >> i, 998244353),
                     m_pow(3, (1004535809 - 1) >> i, 1004535809),
                     m_pow(3, (469762049 - 1) >> i, 469762049)};
    for (long long j = 2; j < (1 << i); j++) w[i][j] = w[i][j - 1] * w[i][1];
  }
  n = 30010;
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % 1000000007;
  inv[n] = m_pow(fac[n], 1000000007 - 2, 1000000007);
  for (long long i = n - 1; i >= 0; i--)
    inv[i] = inv[i + 1] * (i + 1) % 1000000007;
}
inline void change(long long len) {
  for (long long i = 0; i < len; i++) {
    rev[i] = rev[i >> 1] >> 1;
    if (i & 1) rev[i] |= len >> 1;
  }
}
inline void DFT(vector<mint> &y, long long len, long long v) {
  for (long long i = 0; i < len; i++)
    if (i < rev[i]) swap(y[i], y[rev[i]]);
  for (long long i = 2, d = 1; i <= len; i <<= 1, d++) {
    for (long long j = 0; j < len; j += i) {
      mint *W = &w[d][0], a, b;
      for (long long k = j; k < j + i / 2; k++) {
        a = y[k], b = (*W++) * y[k + i / 2];
        y[k] = a + b;
        y[k + i / 2] = a - b;
      }
    }
  }
  if (v == -1) {
    reverse(y.begin() + 1, y.end());
    mint invlen = (mint){1, 1, 1} / len;
    for (long long i = 0; i < (long long)y.size(); i++) y[i] = y[i] * invlen;
  }
}
vector<mint> operator*(vector<mint> a, vector<mint> b) {
  long long n = (long long)a.size(), m = (long long)b.size();
  len = 1;
  while (len < n + m) len <<= 1;
  change(len);
  a.resize(len);
  b.resize(len);
  DFT(a, len, 1);
  DFT(b, len, 1);
  for (long long i = 0; i < len; i++) a[i] = a[i] * b[i];
  DFT(a, len, -1);
  a.resize(n + m - 1);
  for (long long i = 0; i < (long long)a.size(); i++) {
    long long now = a[i].get();
    a[i] = (mint){now, now, now};
  }
  return a;
}
vector<mint> t;
vector<mint> solve(long long n) {
  if (n == 0) {
    return vector<mint>(1, (mint){1, 1, 1});
  }
  vector<mint> a = solve(n >> 1), b = a;
  for (long long i = 0; i < (long long)b.size(); i++)
    b[i] = m_pow(2, (n >> 1) * i, 1000000007) * b[i].a % 1000000007;
  a = a * b;
  if ((long long)a.size() > k + 1) a.resize(k + 1);
  if (n & 1) {
    for (long long i = 0; i < (long long)a.size(); i++)
      a[i] = m_pow(2, i, 1000000007) * a[i].a % 1000000007;
    a = a * t;
    if ((long long)a.size() > k + 1) a.resize(k + 1);
  }
  return a;
}
signed main() {
  prework();
  n = read();
  k = read();
  if (n > k) {
    printf("0\n");
    return 0;
  }
  t.resize(k + 1);
  for (long long i = 1; i <= k; i++) t[i] = (mint){inv[i], inv[i], inv[i]};
  vector<mint> c = solve(n);
  long long ans = 0;
  for (long long i = 1; i <= k; i++)
    add(ans, c[i].a * inv[k - i] % 1000000007, 1000000007);
  mul(ans, fac[k], 1000000007);
  printf("%lld\n", ans);
}