#include <bits/stdc++.h>
using namespace std;
long long i, n, fl, ans, t, j, k;
long long f[50][50];
long long dfs(long long lie, long long han, long long z) {
  long long tem = 0;
  if (lie == 0) return 0;
  if (z == 0) return 1;
  if (han == 1) {
    if (z == 0)
      return 1;
    else
      return 0;
  }
  if (1ll << (han - 2) <= lie) {
    (tem += f[han - 1][z]);
    tem = (tem + dfs(lie - (1ll << (han - 2)), han - 1, z - 1));
  } else {
    (tem += dfs(lie, han - 1, z));
  }
  return tem;
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  fl = -1;
  for (i = 0; i <= 50; i++)
    if ((1ll << i) == t) {
      fl = i;
      break;
    }
  if (fl == -1) {
    puts("0");
    return 0;
  }
  for (i = 2;; i <<= 1) {
    j++;
    if (i - 1 >= n) break;
  }
  f[1][0] = 1;
  for (i = 2; i <= j; i++) {
    f[i][0] = 1;
    for (k = 1; k <= i - 1; k++) f[i][k] = (f[i - 1][k] + f[i - 1][k - 1]);
  }
  ans = 0;
  for (i = 1; i < j; i++) ans += f[i][fl];
  ans += dfs(n - (1ll << j - 1) + 1, j, fl);
  if (t == 1) ans = ans - 1ll;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long i, n, fl, ans, t, j, k;
long long f[50][50];
long long dfs(long long lie, long long han, long long z) {
  long long tem = 0;
  if (lie == 0) return 0;
  if (z == 0) return 1;
  if (han == 1) {
    if (z == 0)
      return 1;
    else
      return 0;
  }
  if (1ll << (han - 2) <= lie) {
    (tem += f[han - 1][z]);
    tem = (tem + dfs(lie - (1ll << (han - 2)), han - 1, z - 1));
  } else {
    (tem += dfs(lie, han - 1, z));
  }
  return tem;
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  fl = -1;
  for (i = 0; i <= 50; i++)
    if ((1ll << i) == t) {
      fl = i;
      break;
    }
  if (fl == -1) {
    puts("0");
    return 0;
  }
  for (i = 2;; i <<= 1) {
    j++;
    if (i - 1 >= n) break;
  }
  f[1][0] = 1;
  for (i = 2; i <= j; i++) {
    f[i][0] = 1;
    for (k = 1; k <= i - 1; k++) f[i][k] = (f[i - 1][k] + f[i - 1][k - 1]);
  }
  ans = 0;
  for (i = 1; i < j; i++) ans += f[i][fl];
  ans += dfs(n - (1ll << j - 1) + 1, j, fl);
  if (t == 1) ans = ans - 1ll;
  printf("%I64d\n", ans);
  return 0;
}

```