#include <bits/stdc++.h>
long long tof, nbin;
long long dp[100][100];
int main() {
  long long n, m;
  scanf("%I64d %I64d", &n, &m);
  n++;
  tof = 0;
  for (; (1LL << tof) < m; tof++)
    ;
  if ((1LL << tof) != m) {
    printf("0\n");
    return 0;
  }
  nbin = 0;
  for (; (1LL << nbin) <= n; nbin++) {
  }
  nbin--;
  dp[0][0] = 1;
  for (long long a = 1; a <= nbin; a++) {
    for (long long b = 0; b <= tof; b++) {
      if (b == 0)
        dp[a][b] = 1;
      else
        dp[a][b] = dp[a - 1][b] + dp[a - 1][b - 1];
    }
  }
  long long ans = 0;
  for (long long i = 0; i < nbin; i++) {
    ans += dp[i][tof];
  }
  long long bnow = nbin;
  n -= (1LL << nbin);
  for (;;) {
    if (tof < 0) break;
    if (bnow == 0) {
      if (tof == 0) ans++;
      break;
    }
    if (n < (1LL << (bnow - 1))) {
      bnow--;
    } else {
      ans += dp[bnow - 1][tof];
      tof--;
      n -= (1LL << (bnow - 1));
      bnow--;
    }
  }
  if (m == 1) ans--;
  printf("%I64d\n", ans);
}

### 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>
long long tof, nbin;
long long dp[100][100];
int main() {
  long long n, m;
  scanf("%I64d %I64d", &n, &m);
  n++;
  tof = 0;
  for (; (1LL << tof) < m; tof++)
    ;
  if ((1LL << tof) != m) {
    printf("0\n");
    return 0;
  }
  nbin = 0;
  for (; (1LL << nbin) <= n; nbin++) {
  }
  nbin--;
  dp[0][0] = 1;
  for (long long a = 1; a <= nbin; a++) {
    for (long long b = 0; b <= tof; b++) {
      if (b == 0)
        dp[a][b] = 1;
      else
        dp[a][b] = dp[a - 1][b] + dp[a - 1][b - 1];
    }
  }
  long long ans = 0;
  for (long long i = 0; i < nbin; i++) {
    ans += dp[i][tof];
  }
  long long bnow = nbin;
  n -= (1LL << nbin);
  for (;;) {
    if (tof < 0) break;
    if (bnow == 0) {
      if (tof == 0) ans++;
      break;
    }
    if (n < (1LL << (bnow - 1))) {
      bnow--;
    } else {
      ans += dp[bnow - 1][tof];
      tof--;
      n -= (1LL << (bnow - 1));
      bnow--;
    }
  }
  if (m == 1) ans--;
  printf("%I64d\n", ans);
}

```