#include <bits/stdc++.h>
using namespace std;
long long a[50][50];
vector<int> get(long long n) {
  vector<int> v;
  int res = 0;
  while (n) {
    v.push_back(n % 2);
    res++;
    n /= 2;
  }
  return v;
}
int get_t(long long t) {
  int res = 1;
  while (t != 1) {
    if (t % 2) return -1;
    t /= 2;
    res++;
  }
  return res;
}
long long cnt(int n, int m) {
  if (n < m) return 0;
  if (m <= 0) return 0;
  n--, m--;
  long long res = 1;
  if (n - m < m) m = n - m;
  for (int i = n - m + 1; i <= n; i++) {
    res *= i;
    res /= (i - (n - m));
  }
  return res;
}
int main(void) {
  long long n, t;
  scanf("%lld%lld", &n, &t);
  n++;
  int cool = get_t(t);
  if (cool == -1) {
    printf("0\n");
    return 0;
  }
  vector<int> len_n = get(n);
  long long res = 0;
  for (int i = cool; i < len_n.size(); i++) {
    res += cnt(i, cool);
  }
  cool--;
  for (int i = len_n.size() - 2; i >= 0; i--) {
    if (len_n[i] == 1) {
      res += cnt(i + 1, cool + 1);
      cool--;
    }
  }
  if (cool == 0) res++;
  if (t == 1) res--;
  printf("%lld\n", res);
  return 0;
}

### Prompt

Generate 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>
using namespace std;
long long a[50][50];
vector<int> get(long long n) {
  vector<int> v;
  int res = 0;
  while (n) {
    v.push_back(n % 2);
    res++;
    n /= 2;
  }
  return v;
}
int get_t(long long t) {
  int res = 1;
  while (t != 1) {
    if (t % 2) return -1;
    t /= 2;
    res++;
  }
  return res;
}
long long cnt(int n, int m) {
  if (n < m) return 0;
  if (m <= 0) return 0;
  n--, m--;
  long long res = 1;
  if (n - m < m) m = n - m;
  for (int i = n - m + 1; i <= n; i++) {
    res *= i;
    res /= (i - (n - m));
  }
  return res;
}
int main(void) {
  long long n, t;
  scanf("%lld%lld", &n, &t);
  n++;
  int cool = get_t(t);
  if (cool == -1) {
    printf("0\n");
    return 0;
  }
  vector<int> len_n = get(n);
  long long res = 0;
  for (int i = cool; i < len_n.size(); i++) {
    res += cnt(i, cool);
  }
  cool--;
  for (int i = len_n.size() - 2; i >= 0; i--) {
    if (len_n[i] == 1) {
      res += cnt(i + 1, cool + 1);
      cool--;
    }
  }
  if (cool == 0) res++;
  if (t == 1) res--;
  printf("%lld\n", res);
  return 0;
}

```