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;
void read(int &digit) {
  digit = 0;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  bool type = false;
  if (c == '-') type = true, c = getchar();
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
  if (type == true) digit = -digit;
}
int n, m;
const int mo = 1000000007;
int powmod(int a, int b) {
  int mjy = 1;
  while (b) {
    if (b & 1) mjy = (1LL * mjy * a) % mo;
    a = (1LL * a * a) % mo;
    b >>= 1;
  }
  return mjy;
}
int fac[30010], _fac[30010], po[30010];
int F[20][30010];
int f[30010], g[30010], h[30010];
const long double pi = acos(-1.0);
struct Complex {
  long double r, i;
  Complex(long double _r = 0.0, long double _i = 0.0) { r = _r, i = _i; }
  Complex operator+(Complex b) { return Complex(r + b.r, i + b.i); }
  Complex operator-(Complex b) { return Complex(r - b.r, i - b.i); }
  Complex operator*(Complex b) {
    return Complex(r * b.r - i * b.i, r * b.i + i * b.r);
  }
} a1[30010 << 2], a2[30010 << 2], b1[30010 << 2], b2[30010 << 2],
    c1[30010 << 2], c2[30010 << 2], c3[30010 << 2];
int l, q;
void FFT(Complex *a, int n, int isdft) {
  for (int i = 0; i < n; i++) {
    int x = 0, y = i;
    for (int j = 1; j < q; j++) {
      x |= y & 1;
      x <<= 1;
      y >>= 1;
    }
    x |= y;
    if (x > i) swap(a[x], a[i]);
  }
  for (int i = 2; i <= n; i <<= 1) {
    Complex wn(cos(isdft * 2.0 * pi / i), sin(isdft * 2.0 * pi / i));
    for (int j = 0; j < n; j += i) {
      Complex w(1, 0), u, v;
      for (int k = j; k < i / 2 + j; k++) {
        u = a[k], v = a[k + i / 2] * w;
        a[k] = u + v;
        a[k + i / 2] = u - v;
        w = w * wn;
      }
    }
  }
  if (isdft == -1)
    for (int i = 0; i < n; i++) a[i].r /= n;
}
const int lim = 1 << 15;
void cheng(int *h, int *f, int *g) {
  for (int i = 0; i <= m; i++)
    a1[i] = Complex(1.0 * (f[i] / lim), 0),
    a2[i] = Complex(1.0 * (f[i] % lim), 0);
  for (int i = 0; i <= m; i++)
    b1[i] = Complex(1.0 * (g[i] / lim), 0),
    b2[i] = Complex(1.0 * (g[i] % lim), 0);
  for (int i = m + 1; i < l; i++) a1[i] = a2[i] = b1[i] = b2[i] = Complex(0, 0);
  FFT(a1, l, 1);
  FFT(a2, l, 1);
  FFT(b1, l, 1);
  FFT(b2, l, 1);
  for (int i = 0; i < l; i++)
    c1[i] = a1[i] * b1[i], c2[i] = a1[i] * b2[i] + a2[i] * b1[i],
    c3[i] = a2[i] * b2[i];
  FFT(c1, l, -1);
  FFT(c2, l, -1);
  FFT(c3, l, -1);
  for (int i = 0; i <= m; i++) {
    long long x = (long long)(c3[i].r + 0.5);
    long long y = (long long)(c2[i].r + 0.5);
    long long z = (long long)(c1[i].r + 0.5);
    x %= mo, y %= mo, z %= mo;
    h[i] = (x + (y * lim)) % mo;
    h[i] = (h[i] + (z * lim * lim) % mo) % mo;
  }
}
int mjy[30010];
void doit(int n) {
  for (int i = 1; i <= m; i++) F[0][i] = 1;
  int now = 2;
  for (int i = 1; now <= n; i++, now <<= 1) {
    f[0] = 0;
    g[0] = 0;
    for (int j = 1; j <= m; j++)
      f[j] = ((1LL * F[i - 1][j] * _fac[j]) % mo *
              (long long)powmod(po[j], now >> 1)) %
             mo;
    for (int j = 1; j <= m; j++) g[j] = (1LL * F[i - 1][j] * _fac[j]) % mo;
    cheng(h, f, g);
    for (int j = 1; j <= m; j++) F[i][j] = (1LL * h[j] * fac[j]) % mo;
  }
  mjy[0] = 1;
  int sum = 1;
  for (int i = 0; n; i++) {
    if (n & 1) {
      for (int j = 0; j <= m; j++)
        f[j] =
            ((1LL * mjy[j] * _fac[j]) % mo * (long long)powmod(po[j], po[i])) %
            mo;
      for (int j = 1; j <= m; j++) g[j] = (1LL * F[i][j] * _fac[j]) % mo;
      g[0] = 0;
      cheng(h, f, g);
      for (int j = 0; j <= m; j++) mjy[j] = (1LL * h[j] * fac[j]) % mo;
    }
    n >>= 1;
  }
}
int main() {
  long long N;
  cin >> N;
  read(m);
  if (N > m) {
    printf("0\n");
    return 0;
  }
  n = N;
  l = 1, q = 0;
  while (l < ((m + 1) << 1)) l <<= 1, q++;
  po[0] = 1;
  fac[0] = _fac[0] = 1;
  for (int i = 1; i <= m; i++)
    po[i] = po[i - 1] * 2 % mo, fac[i] = (1LL * fac[i - 1] * i) % mo,
    _fac[i] = powmod(fac[i], mo - 2);
  doit(n);
  long long zxo = 0;
  for (int i = 1; i <= m; i++)
    zxo = (zxo + 1LL * (1LL * mjy[i] * fac[m]) % mo *
                     ((long long)_fac[i] * _fac[m - i] % mo)) %
          mo;
  cout << zxo << endl;
  return 0;
}