#include <bits/stdc++.h>
using namespace std;
long long n, t, now, tmp, i, j, mul;
long long cnt[60][60];
long long ans[60];
int main() {
  cin >> n >> t;
  n++;
  memset(cnt, 0, sizeof(cnt));
  cnt[0][0] = 1;
  for (i = 1; i <= 50; i++) {
    memcpy(cnt[i], cnt[i - 1], sizeof(cnt[i]));
    for (j = 1; j < 60; j++) cnt[i][j] += cnt[i - 1][j - 1];
  }
  tmp = 1;
  now = n;
  memset(ans, 0, sizeof(ans));
  for (i = 0;; i++) {
    if (now >= tmp) {
      now -= tmp;
      for (j = 0; j < 60; j++) ans[j] += cnt[i][j];
    } else {
      mul = 0;
      break;
    }
    tmp *= 2;
  }
  i--;
  tmp /= 2;
  while (now > 0) {
    if (now >= tmp) {
      now -= tmp;
      for (j = mul; j < 60; j++) ans[j] += cnt[i][j - mul];
      mul++;
    }
    i--;
    tmp /= 2;
  }
  tmp = 0;
  while (t % 2 == 0) {
    tmp++;
    t /= 2;
  }
  if (t == 1) {
    if (tmp == 0) ans[tmp]--;
    cout << ans[tmp] << endl;
  } else
    cout << 0 << endl;
}

### Prompt

Your challenge is to write 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;
long long n, t, now, tmp, i, j, mul;
long long cnt[60][60];
long long ans[60];
int main() {
  cin >> n >> t;
  n++;
  memset(cnt, 0, sizeof(cnt));
  cnt[0][0] = 1;
  for (i = 1; i <= 50; i++) {
    memcpy(cnt[i], cnt[i - 1], sizeof(cnt[i]));
    for (j = 1; j < 60; j++) cnt[i][j] += cnt[i - 1][j - 1];
  }
  tmp = 1;
  now = n;
  memset(ans, 0, sizeof(ans));
  for (i = 0;; i++) {
    if (now >= tmp) {
      now -= tmp;
      for (j = 0; j < 60; j++) ans[j] += cnt[i][j];
    } else {
      mul = 0;
      break;
    }
    tmp *= 2;
  }
  i--;
  tmp /= 2;
  while (now > 0) {
    if (now >= tmp) {
      now -= tmp;
      for (j = mul; j < 60; j++) ans[j] += cnt[i][j - mul];
      mul++;
    }
    i--;
    tmp /= 2;
  }
  tmp = 0;
  while (t % 2 == 0) {
    tmp++;
    t /= 2;
  }
  if (t == 1) {
    if (tmp == 0) ans[tmp]--;
    cout << ans[tmp] << endl;
  } else
    cout << 0 << endl;
}

```