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 int N = 1e5 + 5, mod = 1e9 + 7;
struct atom {
  double x, y;
  atom(double _x = 0, double _y = 0) { x = _x, y = _y; }
  atom operator+(const atom &ch) { return atom(x + ch.x, y + ch.y); }
  atom operator-(const atom &ch) { return atom(x - ch.x, y - ch.y); }
  atom operator*(const atom &ch) {
    return atom(x * ch.x - y * ch.y, x * ch.y + y * ch.x);
  }
} A[N], B[N], C[N], D[N], w, w0, tmp, wi[16][65536];
;
long long n;
int M, l, k, b, r[N], f[N], g[N], fac[N], inv[N], ans;
void FFT(atom *A, int f) {
  for (int i = 0; i < M; i++)
    if (r[i] > i) swap(A[r[i]], A[i]);
  int gg = 0;
  for (int i = 1; i < M; i <<= 1) {
    w.x = cos(acos(-1.0) / i), w.y = sin(acos(-1.0) / i) * f;
    for (int j = 0; j < M; j += (i << 1)) {
      for (int k = 0; k < i; k++) {
        w0 = wi[gg][k], w0.y *= f;
        atom x = A[j + k], y = w0 * A[i + j + k];
        A[j + k] = x + y, A[i + j + k] = x - y;
      }
    }
    gg++;
  }
  if (f == -1)
    for (int i = 0; i < M; i++) A[i].x /= M;
}
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  for (int i = 2; i < N; i++)
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
  for (int i = 2; i < N; i++) inv[i] = 1ll * inv[i] * inv[i - 1] % mod;
  for (int i = 2, k = 0; i <= N; i <<= 1, k++)
    for (int j = 0; j < i; j++)
      wi[k][j] =
          atom(cos(j * acos(-1.0) / (i / 2)), sin(j * acos(-1.0) / (i / 2)));
}
void doit(int *a, int *b, int n) {
  memset(A, 0, sizeof(A));
  memset(B, 0, sizeof(B));
  memset(C, 0, sizeof(C));
  memset(D, 0, sizeof(D));
  for (int i = 0; i < M; i++) {
    A[i].x = a[i] >> 15;
    B[i].x = a[i] & 32767;
    C[i].x = b[i] >> 15;
    D[i].x = b[i] & 32767;
  }
  FFT(A, 1), FFT(B, 1), FFT(C, 1), FFT(D, 1);
  for (int i = 0; i < M; i++) {
    tmp = A[i] * D[i] + B[i] * C[i];
    A[i] = A[i] * C[i];
    C[i] = B[i] * D[i];
    B[i] = tmp;
  }
  FFT(A, -1), FFT(B, -1), FFT(C, -1), memset(a, 0, sizeof(a));
  for (int i = 0; i <= n; i++)
    a[i] = ((llround(A[i].x) % mod << 30) + (llround(B[i].x) % mod << 15) +
            llround(C[i].x) % mod) %
           mod;
}
int c[N];
void work(int *a, int *b, int n, int p) {
  int t = 1;
  memset(c, 0, sizeof(c));
  for (int i = 0; i <= n; i++)
    c[i] = 1ll * a[i] * t % mod, t = 1ll * t * p % mod;
  doit(c, b, n);
  memcpy(a, c, sizeof c);
}
int Calc(int n, int m) {
  return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
int main() {
  cin >> n >> k;
  if (n > k) {
    puts("0");
    return 0;
  }
  for (M = 1; M <= k + k; M <<= 1, l++)
    ;
  for (int i = 0; i < M; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  init(), g[0] = 1, b = 2;
  for (int i = 1; i <= k; i++) f[i] = inv[i];
  for (; n; n >>= 1, work(f, f, k, b), b = 1ll * b * b % mod)
    if (n & 1) work(g, f, k, b);
  for (int i = n; i <= k; i++)
    (ans += 1ll * g[i] * fac[i] % mod * Calc(k, i) % mod) %= mod;
  printf("%d\n", ans);
}