#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
long long dp[62][62][2], ans;
inline void add(int i, int j, int e, long long a) {
  if (i < 0) {
    if (j == k) ans += a;
    return;
  }
  dp[i][j][e] += a;
}
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    printf("0\n");
    exit(0);
  }
  int cnt = 0;
  while (t) {
    t /= 2;
    ++cnt;
  }
  k = cnt;
  ++n;
  dp[61][0][0] = 1;
  for (int i = (61), _b = (0); i >= _b; i--) {
    for (int _n((62) - 1), j(0); j <= _n; j++) {
      for (int _n((2) - 1), e(0); e <= _n; e++) {
        if (!dp[i][j][e]) continue;
        int ne = e;
        if (n & 1LL << i) ne = 1;
        add(i - 1, j, ne, dp[i][j][e]);
        if ((n & 1LL << i) || e) {
          add(i - 1, j + 1, e, dp[i][j][e]);
        }
      }
    }
  }
  if (k == 1) --ans;
  cout << ans << 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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
long long dp[62][62][2], ans;
inline void add(int i, int j, int e, long long a) {
  if (i < 0) {
    if (j == k) ans += a;
    return;
  }
  dp[i][j][e] += a;
}
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    printf("0\n");
    exit(0);
  }
  int cnt = 0;
  while (t) {
    t /= 2;
    ++cnt;
  }
  k = cnt;
  ++n;
  dp[61][0][0] = 1;
  for (int i = (61), _b = (0); i >= _b; i--) {
    for (int _n((62) - 1), j(0); j <= _n; j++) {
      for (int _n((2) - 1), e(0); e <= _n; e++) {
        if (!dp[i][j][e]) continue;
        int ne = e;
        if (n & 1LL << i) ne = 1;
        add(i - 1, j, ne, dp[i][j][e]);
        if ((n & 1LL << i) || e) {
          add(i - 1, j + 1, e, dp[i][j][e]);
        }
      }
    }
  }
  if (k == 1) --ans;
  cout << ans << endl;
}

```