#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = (int)1e9 + 7;
const int MAXN = (int)2e6 + 7;
long long n, t;
long long second[MAXN];
long long c[55][55];
long long all(long long len, long long cnt) {
  if (cnt < 0 || cnt > len) return 0;
  return c[len][cnt];
}
long long calculate(long long cnt, long long limit) {
  vector<long long> d;
  while (limit) {
    d.push_back(limit & 1);
    limit >>= 1;
  }
  reverse(d.begin(), d.end());
  long long prefix = 0;
  long long res = 0;
  for (long long i = 0; i < d.size(); i++) {
    for (long long put = 0; put < d[i]; put++) {
      res += all(d.size() - i - 1, cnt - prefix);
    }
    prefix += d[i];
  }
  return res + (prefix == cnt);
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  cin >> n >> t;
  long long tt = t;
  if (t & (t - 1)) {
    cout << "0";
    return 0;
  }
  long long highest = 0;
  while (t % 2 == 0) {
    t >>= 1;
    highest++;
  }
  highest++;
  long long res = calculate(highest, n + 1);
  for (int i = 0; i < 2; i++)
    if (__builtin_popcount(i) == highest) res--;
  cout << res;
  return 0;
}

### Prompt

Your task is to create 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;
const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = (int)1e9 + 7;
const int MAXN = (int)2e6 + 7;
long long n, t;
long long second[MAXN];
long long c[55][55];
long long all(long long len, long long cnt) {
  if (cnt < 0 || cnt > len) return 0;
  return c[len][cnt];
}
long long calculate(long long cnt, long long limit) {
  vector<long long> d;
  while (limit) {
    d.push_back(limit & 1);
    limit >>= 1;
  }
  reverse(d.begin(), d.end());
  long long prefix = 0;
  long long res = 0;
  for (long long i = 0; i < d.size(); i++) {
    for (long long put = 0; put < d[i]; put++) {
      res += all(d.size() - i - 1, cnt - prefix);
    }
    prefix += d[i];
  }
  return res + (prefix == cnt);
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  cin >> n >> t;
  long long tt = t;
  if (t & (t - 1)) {
    cout << "0";
    return 0;
  }
  long long highest = 0;
  while (t % 2 == 0) {
    t >>= 1;
    highest++;
  }
  highest++;
  long long res = calculate(highest, n + 1);
  for (int i = 0; i < 2; i++)
    if (__builtin_popcount(i) == highest) res--;
  cout << res;
  return 0;
}

```