#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (((a * b) / gcd(a, b))); }
long long phi(long long n) {
  long long result = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
void nxi(long long& n) {
  bool min = 0;
  char c;
  n = 0;
  while ((c = getc(stdin)) && c <= 32)
    ;
  if (c == '-')
    min = 1;
  else
    n = c - 48;
  while ((c = getc(stdin)) && c > 32) n = (n << 3) + (n << 1) + c - 48;
  if (min) n = -n;
}
void prl(long long n) {
  if (n == 0) {
    puts("0");
    return;
  }
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  static long long s[10];
  long long top = 0;
  while (n > 0) s[top++] = n % 10, n /= 10;
  while (top--) putchar(s[top] + 48);
  puts("");
}
using namespace std;
const long long nmax = 2100;
long long n, t, cb[nmax][nmax], e, maxbit = 62;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> t;
  if (t & (t - 1)) return cout << 0 << endl, 0;
  int e = 0;
  for (; t != 1; t >>= 1, ++e)
    ;
  for (int i = 0; i < maxbit; ++i)
    for (int j = 0; j <= i; ++j)
      cb[i][j] = j ? cb[i - 1][j - 1] + cb[i - 1][j] : 1;
  long long ans = 0;
  int x = 0;
  n += 2;
  ++e;
  for (int b = maxbit - 1; b >= 0; --b) {
    if (n & (1ll << b)) {
      ans += cb[b][e - x];
      ++x;
      if (x > e) break;
    }
  }
  if (e == 1) ans -= 1;
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):

<image>

Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.

Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".

Input

A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

In a single line print a single integer — the answer to the problem. 

Examples

Input

1 1


Output

1


Input

3 2


Output

1


Input

3 3


Output

0


Input

1000000000000 1048576


Output

118606527258

### Response

```cpp
#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (((a * b) / gcd(a, b))); }
long long phi(long long n) {
  long long result = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
void nxi(long long& n) {
  bool min = 0;
  char c;
  n = 0;
  while ((c = getc(stdin)) && c <= 32)
    ;
  if (c == '-')
    min = 1;
  else
    n = c - 48;
  while ((c = getc(stdin)) && c > 32) n = (n << 3) + (n << 1) + c - 48;
  if (min) n = -n;
}
void prl(long long n) {
  if (n == 0) {
    puts("0");
    return;
  }
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  static long long s[10];
  long long top = 0;
  while (n > 0) s[top++] = n % 10, n /= 10;
  while (top--) putchar(s[top] + 48);
  puts("");
}
using namespace std;
const long long nmax = 2100;
long long n, t, cb[nmax][nmax], e, maxbit = 62;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> t;
  if (t & (t - 1)) return cout << 0 << endl, 0;
  int e = 0;
  for (; t != 1; t >>= 1, ++e)
    ;
  for (int i = 0; i < maxbit; ++i)
    for (int j = 0; j <= i; ++j)
      cb[i][j] = j ? cb[i - 1][j - 1] + cb[i - 1][j] : 1;
  long long ans = 0;
  int x = 0;
  n += 2;
  ++e;
  for (int b = maxbit - 1; b >= 0; --b) {
    if (n & (1ll << b)) {
      ans += cb[b][e - x];
      ++x;
      if (x > e) break;
    }
  }
  if (e == 1) ans -= 1;
  cout << ans << endl;
  return 0;
}

```