#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int K = 32768;
const int K2 = K * 2;
const long double PI = acos((long double)-1.0);
struct DP {
  int v[K], n;
  DP() { memset(v, 0, sizeof(v)); }
};
int fact[K];
void FFT(complex<long double> a[], int n, int oper) {
  for (int i = 1, j = 0; i < n; i++) {
    for (int s = n; j ^= s >>= 1, ~j & s;)
      ;
    if (i < j) swap(a[i], a[j]);
  }
  for (int m = 1; m < n; m *= 2) {
    long double p = PI / m * oper;
    complex<long double> w = complex<long double>(cos(p), sin(p));
    for (int i = 0; i < n; i += m * 2) {
      complex<long double> unit = 1;
      for (int j = 0; j < m; j++) {
        complex<long double> &x = a[i + j + m], &y = a[i + j], t = unit * x;
        x = y - t;
        y = y + t;
        unit *= w;
      }
    }
  }
  if (oper == -1)
    for (int i = 0; i < n; i++) a[i] /= n;
}
int power(int x, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans = (long long)ans * x % mod;
    }
    x = (long long)x * x % mod;
    n >>= 1;
  }
  return ans;
}
DP operator*(DP a, DP b) {
  for (int i = 0; i < K; i++) {
    a.v[i] = (long long)a.v[i] * power(2, b.n * i) % mod;
  }
  const int base = sqrt(mod);
  static complex<long double> a0[K2], a1[K2], b0[K2], b1[K2];
  static complex<long double> c00[K2], c01[K2], c11[K2];
  for (int i = 0; i < K2; i++) {
    a0[i] = a1[i] = b0[i] = b1[i] = 0;
  }
  for (int i = 0; i < K; i++) {
    a0[i] = a.v[i] % base;
    a1[i] = a.v[i] / base;
    b0[i] = b.v[i] % base;
    b1[i] = b.v[i] / base;
  }
  FFT(a0, K2, 1);
  FFT(a1, K2, 1);
  FFT(b0, K2, 1);
  FFT(b1, K2, 1);
  for (int i = 0; i < K2; i++) {
    c00[i] = a0[i] * b0[i];
    c01[i] = (a0[i] + a1[i]) * (b0[i] + b1[i]);
    c11[i] = a1[i] * b1[i];
  }
  FFT(c00, K2, -1);
  FFT(c01, K2, -1);
  FFT(c11, K2, -1);
  DP ans;
  ans.n = a.n + b.n;
  for (int i = 0; i < K; i++) {
    long long k00 = (long long)(real(c00[i]) + 0.5) % mod;
    long long k01 = (long long)(real(c01[i]) + 0.5) % mod;
    long long k11 = (long long)(real(c11[i]) + 0.5) % mod;
    ans.v[i] =
        ((k11 * base + (k01 - k00 - k11) % mod + mod) % mod * base + k00) % mod;
  }
  return ans;
}
DP cal(int n) {
  DP x, ans;
  ans.n = 0;
  ans.v[0] = 1;
  x.n = 1;
  for (int i = 1; i < K; i++) {
    x.v[i] = power(fact[i], mod - 2);
  }
  while (n) {
    if (n & 1) {
      ans = ans * x;
    }
    x = x * x;
    n >>= 1;
  }
  return ans;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < K; i++) {
    fact[i] = (long long)fact[i - 1] * i % mod;
  }
  long long n;
  int k;
  cin >> n >> k;
  if (n > k) {
    cout << 0 << endl;
    return 0;
  }
  DP dp = cal(n);
  int ans = 0;
  int c = 1;
  for (int i = 0; i <= k; i++) {
    ans = (ans + (long long)dp.v[i] * fact[i] % mod * c) % mod;
    c = (long long)c * (k - i) % mod * power(i + 1, mod - 2) % mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int K = 32768;
const int K2 = K * 2;
const long double PI = acos((long double)-1.0);
struct DP {
  int v[K], n;
  DP() { memset(v, 0, sizeof(v)); }
};
int fact[K];
void FFT(complex<long double> a[], int n, int oper) {
  for (int i = 1, j = 0; i < n; i++) {
    for (int s = n; j ^= s >>= 1, ~j & s;)
      ;
    if (i < j) swap(a[i], a[j]);
  }
  for (int m = 1; m < n; m *= 2) {
    long double p = PI / m * oper;
    complex<long double> w = complex<long double>(cos(p), sin(p));
    for (int i = 0; i < n; i += m * 2) {
      complex<long double> unit = 1;
      for (int j = 0; j < m; j++) {
        complex<long double> &x = a[i + j + m], &y = a[i + j], t = unit * x;
        x = y - t;
        y = y + t;
        unit *= w;
      }
    }
  }
  if (oper == -1)
    for (int i = 0; i < n; i++) a[i] /= n;
}
int power(int x, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans = (long long)ans * x % mod;
    }
    x = (long long)x * x % mod;
    n >>= 1;
  }
  return ans;
}
DP operator*(DP a, DP b) {
  for (int i = 0; i < K; i++) {
    a.v[i] = (long long)a.v[i] * power(2, b.n * i) % mod;
  }
  const int base = sqrt(mod);
  static complex<long double> a0[K2], a1[K2], b0[K2], b1[K2];
  static complex<long double> c00[K2], c01[K2], c11[K2];
  for (int i = 0; i < K2; i++) {
    a0[i] = a1[i] = b0[i] = b1[i] = 0;
  }
  for (int i = 0; i < K; i++) {
    a0[i] = a.v[i] % base;
    a1[i] = a.v[i] / base;
    b0[i] = b.v[i] % base;
    b1[i] = b.v[i] / base;
  }
  FFT(a0, K2, 1);
  FFT(a1, K2, 1);
  FFT(b0, K2, 1);
  FFT(b1, K2, 1);
  for (int i = 0; i < K2; i++) {
    c00[i] = a0[i] * b0[i];
    c01[i] = (a0[i] + a1[i]) * (b0[i] + b1[i]);
    c11[i] = a1[i] * b1[i];
  }
  FFT(c00, K2, -1);
  FFT(c01, K2, -1);
  FFT(c11, K2, -1);
  DP ans;
  ans.n = a.n + b.n;
  for (int i = 0; i < K; i++) {
    long long k00 = (long long)(real(c00[i]) + 0.5) % mod;
    long long k01 = (long long)(real(c01[i]) + 0.5) % mod;
    long long k11 = (long long)(real(c11[i]) + 0.5) % mod;
    ans.v[i] =
        ((k11 * base + (k01 - k00 - k11) % mod + mod) % mod * base + k00) % mod;
  }
  return ans;
}
DP cal(int n) {
  DP x, ans;
  ans.n = 0;
  ans.v[0] = 1;
  x.n = 1;
  for (int i = 1; i < K; i++) {
    x.v[i] = power(fact[i], mod - 2);
  }
  while (n) {
    if (n & 1) {
      ans = ans * x;
    }
    x = x * x;
    n >>= 1;
  }
  return ans;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < K; i++) {
    fact[i] = (long long)fact[i - 1] * i % mod;
  }
  long long n;
  int k;
  cin >> n >> k;
  if (n > k) {
    cout << 0 << endl;
    return 0;
  }
  DP dp = cal(n);
  int ans = 0;
  int c = 1;
  for (int i = 0; i <= k; i++) {
    ans = (ans + (long long)dp.v[i] * fact[i] % mod * c) % mod;
    c = (long long)c * (k - i) % mod * power(i + 1, mod - 2) % mod;
  }
  cout << ans << endl;
  return 0;
}

```