#include <bits/stdc++.h>
using namespace std;
const long long N = (long long)1e12 + 100;
const int MAX = 64;
int max_pow = 0;
long long n, t, dp[MAX][MAX];
bool used[MAX];
void calc(int val) {
  max_pow = max(max_pow, val);
  if (used[val]) return;
  used[val] = true;
  if (val == 0) {
    dp[0][0] = 1;
    return;
  }
  calc(val - 1);
  for (int i = 0; i <= val; i++) {
    dp[val][i] = dp[val - 1][i];
    if (i) dp[val][i] += dp[val - 1][i - 1];
  }
}
long long solve(long long n, int t) {
  assert(n >= 0);
  if (!n) return 0;
  if (t < 0) return 0;
  long long res = 0;
  int cur_pow = 0;
  long long cur_h = 1, h = 1;
  while (h <= n) {
    calc(cur_pow);
    res += dp[cur_pow][t];
    if (h + 2 * cur_h <= n) {
      cur_h *= 2;
      h += cur_h;
      cur_pow++;
    } else
      break;
  }
  long long tail = n - h, max_h = 1;
  int max_can = 0;
  if (tail == 0) return res;
  while (max_h <= tail) {
    max_h *= 2;
    max_can++;
  }
  while (t >= 0 && max_can >= 0) {
    while (tail < max_h) max_can--, max_h /= 2;
    res += dp[max_can][t];
    t--;
    tail -= max_h;
  }
  return res;
}
int main() {
  cin >> n >> t;
  long long tmp = t;
  int p = 0;
  while (tmp % 2 == 0) tmp /= 2, p++;
  if (tmp != 1) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = solve(n + 1, p);
  if (p)
    cout << ans << endl;
  else
    cout << ans - 1 << 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 long long N = (long long)1e12 + 100;
const int MAX = 64;
int max_pow = 0;
long long n, t, dp[MAX][MAX];
bool used[MAX];
void calc(int val) {
  max_pow = max(max_pow, val);
  if (used[val]) return;
  used[val] = true;
  if (val == 0) {
    dp[0][0] = 1;
    return;
  }
  calc(val - 1);
  for (int i = 0; i <= val; i++) {
    dp[val][i] = dp[val - 1][i];
    if (i) dp[val][i] += dp[val - 1][i - 1];
  }
}
long long solve(long long n, int t) {
  assert(n >= 0);
  if (!n) return 0;
  if (t < 0) return 0;
  long long res = 0;
  int cur_pow = 0;
  long long cur_h = 1, h = 1;
  while (h <= n) {
    calc(cur_pow);
    res += dp[cur_pow][t];
    if (h + 2 * cur_h <= n) {
      cur_h *= 2;
      h += cur_h;
      cur_pow++;
    } else
      break;
  }
  long long tail = n - h, max_h = 1;
  int max_can = 0;
  if (tail == 0) return res;
  while (max_h <= tail) {
    max_h *= 2;
    max_can++;
  }
  while (t >= 0 && max_can >= 0) {
    while (tail < max_h) max_can--, max_h /= 2;
    res += dp[max_can][t];
    t--;
    tail -= max_h;
  }
  return res;
}
int main() {
  cin >> n >> t;
  long long tmp = t;
  int p = 0;
  while (tmp % 2 == 0) tmp /= 2, p++;
  if (tmp != 1) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = solve(n + 1, p);
  if (p)
    cout << ans << endl;
  else
    cout << ans - 1 << endl;
  return 0;
}

```