#include <bits/stdc++.h>
using namespace std;
long long f[50][50];
long long get(long long x, long long y) {
  if (y == -1)
    return 1LL;
  else
    return f[x][y];
}
long long gao(long long mm, long long mt, long long x, long long t) {
  if (mt == 0) return (t == -1);
  if (t < -1) return 0;
  if (x < mm / 2LL)
    return gao(mm / 2LL, mt - 1, x, t);
  else {
    return gao(mm / 2LL, mt - 1, x - mm / 2LL, t - 1) + get(mt - 1, t);
  }
}
int main() {
  f[1][0] = 1;
  for (long long i = 2; i < 50; ++i) {
    for (long long j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (j) f[i][j] += f[i - 1][j - 1];
    }
    f[i][0]++;
  }
  long long n, t;
  scanf("%I64d%I64d", &n, &t);
  ++n;
  if ((t & -t) != t) {
    printf("0\n");
    return 0;
  }
  long long ss = 1, st = 0;
  while (ss <= n) {
    ss = ss * 2LL;
    ++st;
  }
  long long rt = 0;
  while (t > 1) {
    rt++;
    t >>= 1LL;
  }
  long long ans = gao(ss, st, n, rt);
  if (rt == 0) ans--;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 f[50][50];
long long get(long long x, long long y) {
  if (y == -1)
    return 1LL;
  else
    return f[x][y];
}
long long gao(long long mm, long long mt, long long x, long long t) {
  if (mt == 0) return (t == -1);
  if (t < -1) return 0;
  if (x < mm / 2LL)
    return gao(mm / 2LL, mt - 1, x, t);
  else {
    return gao(mm / 2LL, mt - 1, x - mm / 2LL, t - 1) + get(mt - 1, t);
  }
}
int main() {
  f[1][0] = 1;
  for (long long i = 2; i < 50; ++i) {
    for (long long j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (j) f[i][j] += f[i - 1][j - 1];
    }
    f[i][0]++;
  }
  long long n, t;
  scanf("%I64d%I64d", &n, &t);
  ++n;
  if ((t & -t) != t) {
    printf("0\n");
    return 0;
  }
  long long ss = 1, st = 0;
  while (ss <= n) {
    ss = ss * 2LL;
    ++st;
  }
  long long rt = 0;
  while (t > 1) {
    rt++;
    t >>= 1LL;
  }
  long long ans = gao(ss, st, n, rt);
  if (rt == 0) ans--;
  printf("%I64d\n", ans);
  return 0;
}

```