#include <bits/stdc++.h>
using namespace std;
const int N = 60;
long long n, t, C[N][N];
long long dfs(long long n, int t) {
  if (n == 0 || t < 0) return 0;
  for (int i = N - 1; i >= 0; --i)
    if ((1ll << i) <= n) return C[i][t] + dfs(n - (1ll << i), t - 1);
  return 0;
}
long long solve(long long n, int x) {
  long long ret = 0;
  for (int i = (0); i < (N); ++i)
    if ((1ll << i) <= n) {
      ret += C[i][x];
      n -= 1ll << i;
    } else {
      ret += dfs(n, x);
      break;
    }
  return ret;
}
int main() {
  for (int i = (0); i < (N); ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = (1); j < (i); ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  scanf("%lld%lld", &n, &t);
  if (t == (t & -t)) {
    int x = 0;
    while ((1ll << x) != t) ++x;
    long long ans = solve(n + 1, x) - (x == 0);
    printf("%lld", ans);
  } else {
    printf("0");
  }
  return 0;
}

### Prompt

Your task is to create 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 N = 60;
long long n, t, C[N][N];
long long dfs(long long n, int t) {
  if (n == 0 || t < 0) return 0;
  for (int i = N - 1; i >= 0; --i)
    if ((1ll << i) <= n) return C[i][t] + dfs(n - (1ll << i), t - 1);
  return 0;
}
long long solve(long long n, int x) {
  long long ret = 0;
  for (int i = (0); i < (N); ++i)
    if ((1ll << i) <= n) {
      ret += C[i][x];
      n -= 1ll << i;
    } else {
      ret += dfs(n, x);
      break;
    }
  return ret;
}
int main() {
  for (int i = (0); i < (N); ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = (1); j < (i); ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  scanf("%lld%lld", &n, &t);
  if (t == (t & -t)) {
    int x = 0;
    while ((1ll << x) != t) ++x;
    long long ans = solve(n + 1, x) - (x == 0);
    printf("%lld", ans);
  } else {
    printf("0");
  }
  return 0;
}

```