#include <bits/stdc++.h>
using namespace std;
template <class T, class L>
inline bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
inline bool smin(T &x, L y) {
  return y < x ? (x = y, 1) : 0;
}
const int maxn = 50;
long long c[maxn][maxn], n, t, ans;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> t, n++;
  if (__builtin_popcountll(t) != 1) return cout << "0\n", 0;
  t = __builtin_ctzll(t) + 1;
  for (int i = c[0][0] = 1; i < maxn; i++)
    for (int j = c[i][0] = 1; j <= i; j++)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  ;
  ans = (__builtin_popcountll(n) == t) - (t == 1);
  for (int i = maxn; t >= 0 && ~i; i--)
    if (n >> i & 1) ans += c[i][t--];
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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;
template <class T, class L>
inline bool smax(T &x, L y) {
  return x < y ? (x = y, 1) : 0;
}
template <class T, class L>
inline bool smin(T &x, L y) {
  return y < x ? (x = y, 1) : 0;
}
const int maxn = 50;
long long c[maxn][maxn], n, t, ans;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> t, n++;
  if (__builtin_popcountll(t) != 1) return cout << "0\n", 0;
  t = __builtin_ctzll(t) + 1;
  for (int i = c[0][0] = 1; i < maxn; i++)
    for (int j = c[i][0] = 1; j <= i; j++)
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  ;
  ans = (__builtin_popcountll(n) == t) - (t == 1);
  for (int i = maxn; t >= 0 && ~i; i--)
    if (n >> i & 1) ans += c[i][t--];
  cout << ans << '\n';
  return 0;
}

```