#include <bits/stdc++.h>
using namespace std;
long long c[100][100];
void prepare() {
  int i, j;
  memset(c, 0, sizeof(c));
  c[1][0] = c[1][1] = 1;
  for (i = 2; i < 50; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
}
long long calc(long long n, int p) {
  if (n == 0 || p < 0) return 0;
  if (n == 1) return p == 0;
  int i;
  for (i = 1; (1LL << i) <= n; i++)
    ;
  i--;
  return c[i][p] + calc(n - (1LL << i), p - 1);
}
int main() {
  prepare();
  long long n, t, s, ans;
  int i, p = 0;
  scanf("%I64d%I64d", &n, &t);
  while (t % 2 == 0) {
    t /= 2;
    p++;
  }
  if (t != 1) {
    puts("0");
    return 0;
  }
  for (s = ans = 0, i = 1; s + (1LL << i) <= n; i++) {
    s += 1LL << i;
    ans += c[i][p];
  }
  n -= s;
  ans += calc(n, p);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 c[100][100];
void prepare() {
  int i, j;
  memset(c, 0, sizeof(c));
  c[1][0] = c[1][1] = 1;
  for (i = 2; i < 50; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
}
long long calc(long long n, int p) {
  if (n == 0 || p < 0) return 0;
  if (n == 1) return p == 0;
  int i;
  for (i = 1; (1LL << i) <= n; i++)
    ;
  i--;
  return c[i][p] + calc(n - (1LL << i), p - 1);
}
int main() {
  prepare();
  long long n, t, s, ans;
  int i, p = 0;
  scanf("%I64d%I64d", &n, &t);
  while (t % 2 == 0) {
    t /= 2;
    p++;
  }
  if (t != 1) {
    puts("0");
    return 0;
  }
  for (s = ans = 0, i = 1; s + (1LL << i) <= n; i++) {
    s += 1LL << i;
    ans += c[i][p];
  }
  n -= s;
  ans += calc(n, p);
  printf("%I64d\n", ans);
  return 0;
}

```