#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int prime = 9241;
const long double pi = acos(-1.);
long long dp[55][55];
long long rec(int i, int j, long long ub, int f) {
  long long res = -1ll;
  if (ub < 0ll)
    res = 0ll;
  else if (ub >= (1ll << i))
    res = ((i >= 0 && f - j >= 0) ? dp[i][f - j] : 0ll);
  else
    res = rec(i - 1, j, ub, f) + rec(i - 1, j + 1, ub - (1ll << (i - 1)), f);
  return res;
}
int main() {
  long long n, t;
  cin >> n >> t;
  int logt = -1;
  for (int j = 0; j < 50; j++)
    if ((1ll << j) == t) logt = j;
  if (logt == -1) {
    cout << 0 << endl;
    return 0;
  }
  dp[0][0] = 1;
  for (int i = 1; i < 55; i++)
    for (int j = 0; j < 55; j++)
      dp[i][j] = dp[i - 1][j] + (j > 0 ? dp[i - 1][j - 1] : 0ll);
  int p = 1;
  long long ans = 0ll;
  while (n >= (1ll << p)) {
    if (p != 0) ans += dp[p][logt];
    n -= 1ll << p;
    p++;
  }
  cout << ans + rec(50, 0, n, logt) << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 int INF = 1000000000;
const int prime = 9241;
const long double pi = acos(-1.);
long long dp[55][55];
long long rec(int i, int j, long long ub, int f) {
  long long res = -1ll;
  if (ub < 0ll)
    res = 0ll;
  else if (ub >= (1ll << i))
    res = ((i >= 0 && f - j >= 0) ? dp[i][f - j] : 0ll);
  else
    res = rec(i - 1, j, ub, f) + rec(i - 1, j + 1, ub - (1ll << (i - 1)), f);
  return res;
}
int main() {
  long long n, t;
  cin >> n >> t;
  int logt = -1;
  for (int j = 0; j < 50; j++)
    if ((1ll << j) == t) logt = j;
  if (logt == -1) {
    cout << 0 << endl;
    return 0;
  }
  dp[0][0] = 1;
  for (int i = 1; i < 55; i++)
    for (int j = 0; j < 55; j++)
      dp[i][j] = dp[i - 1][j] + (j > 0 ? dp[i - 1][j - 1] : 0ll);
  int p = 1;
  long long ans = 0ll;
  while (n >= (1ll << p)) {
    if (p != 0) ans += dp[p][logt];
    n -= 1ll << p;
    p++;
  }
  cout << ans + rec(50, 0, n, logt) << endl;
  return 0;
}

```