#include <bits/stdc++.h>
using namespace std;
const int mod = 9901;
template <class T>
int countbit(T n) {
  int t = 0;
  while (n) n &= n - 1, ++t;
  return t;
}
template <class T>
T lowbit(T n) {
  return n & (-n);
}
const double eps = 1e-8;
const double pi = acos(-1.0);
const int INF = 0x7f7f7f7f;
const int maxn = 111111;
void File() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
long long c[66][66];
long long go(long long n, long long t) {
  int cnt = 0;
  long long tt = t;
  while (tt) tt >>= 1, ++cnt;
  int arr[65];
  int i, size = 0;
  while (n) arr[size++] = n & 1, n >>= 1;
  reverse(arr, arr + size);
  long long ans = 0;
  bool flag;
  int ct = count(arr, arr + size, 1);
  for (i = 0; i < size; ++i) {
    flag = true;
    if (arr[i]) {
      if (size - i == ct && ct == cnt && cnt)
        ans++, flag = false;
      else {
        if (cnt >= 0) ans += c[size - i - 1][cnt], --cnt;
      }
      --ct;
    }
    if (i && !cnt) {
      ans++;
      break;
    }
    if (!flag) break;
  }
  if (t == 1) ans--;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int i, j;
  for (i = 0; i < 66; ++i) {
    c[i][i] = c[i][0] = 1;
  }
  for (i = 1; i < 66; ++i) {
    for (j = 1; j < i; ++j) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
  long long n, t;
  cin >> n >> t;
  int k = countbit(t);
  if (k == 1)
    cout << go(n + 1, t) << endl;
  else
    cout << 0 << endl;
  return 0;
}

### Prompt

Generate 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;
const int mod = 9901;
template <class T>
int countbit(T n) {
  int t = 0;
  while (n) n &= n - 1, ++t;
  return t;
}
template <class T>
T lowbit(T n) {
  return n & (-n);
}
const double eps = 1e-8;
const double pi = acos(-1.0);
const int INF = 0x7f7f7f7f;
const int maxn = 111111;
void File() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
long long c[66][66];
long long go(long long n, long long t) {
  int cnt = 0;
  long long tt = t;
  while (tt) tt >>= 1, ++cnt;
  int arr[65];
  int i, size = 0;
  while (n) arr[size++] = n & 1, n >>= 1;
  reverse(arr, arr + size);
  long long ans = 0;
  bool flag;
  int ct = count(arr, arr + size, 1);
  for (i = 0; i < size; ++i) {
    flag = true;
    if (arr[i]) {
      if (size - i == ct && ct == cnt && cnt)
        ans++, flag = false;
      else {
        if (cnt >= 0) ans += c[size - i - 1][cnt], --cnt;
      }
      --ct;
    }
    if (i && !cnt) {
      ans++;
      break;
    }
    if (!flag) break;
  }
  if (t == 1) ans--;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int i, j;
  for (i = 0; i < 66; ++i) {
    c[i][i] = c[i][0] = 1;
  }
  for (i = 1; i < 66; ++i) {
    for (j = 1; j < i; ++j) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
  long long n, t;
  cin >> n >> t;
  int k = countbit(t);
  if (k == 1)
    cout << go(n + 1, t) << endl;
  else
    cout << 0 << endl;
  return 0;
}

```