#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double EPS = 1e-9, INF = 1e12;
int m, d[70];
long long dp[70][70][2];
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  for (; n; n /= 2) d[m++] = n % 2;
  reverse(d, d + m);
  int bit = __builtin_ctzll(t) + 1;
  if (__builtin_popcountll(t) != 1) {
    cout << 0 << endl;
    return 0;
  }
  dp[0][0][0] = 1;
  for (int i = 0; i < (int)m; i++)
    for (int j = 0; j < (int)i + 1; j++)
      for (int k = 0; k < (int)2; k++)
        if (dp[i][j][k]) {
          for (int l = 0; l < (int)2; l++) {
            int nk = k || l < d[i];
            if (!nk && l > d[i]) continue;
            dp[i + 1][j + l][nk] += dp[i][j][k];
          }
        }
  cout << dp[m][bit][0] + dp[m][bit][1] - (t == 1) << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 int inf = (int)1e9;
const double EPS = 1e-9, INF = 1e12;
int m, d[70];
long long dp[70][70][2];
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  for (; n; n /= 2) d[m++] = n % 2;
  reverse(d, d + m);
  int bit = __builtin_ctzll(t) + 1;
  if (__builtin_popcountll(t) != 1) {
    cout << 0 << endl;
    return 0;
  }
  dp[0][0][0] = 1;
  for (int i = 0; i < (int)m; i++)
    for (int j = 0; j < (int)i + 1; j++)
      for (int k = 0; k < (int)2; k++)
        if (dp[i][j][k]) {
          for (int l = 0; l < (int)2; l++) {
            int nk = k || l < d[i];
            if (!nk && l > d[i]) continue;
            dp[i + 1][j + l][nk] += dp[i][j][k];
          }
        }
  cout << dp[m][bit][0] + dp[m][bit][1] - (t == 1) << endl;
  return 0;
}

```