#include <bits/stdc++.h>
long long cnt[100][100];
long long n, t, k;
long long go(long long rest, long long level, long long x) {
  if (level == 1) {
    if (rest == 1)
      return x == 0;
    else
      return x == 1;
  }
  if (rest > (1ll << (level - 1)))
    return go(rest - (1ll << (level - 1)), level - 1, x - 1) +
           cnt[level - 1][x];
  else
    return go(rest, level - 1, x);
}
int main() {
  cnt[1][0] = cnt[1][1] = 1;
  for (int i = 2; i <= 100; ++i) {
    for (int j = 0; j <= i; ++j) {
      cnt[i][j] += cnt[i - 1][j];
      if (j > 0) cnt[i][j] += cnt[i - 1][j - 1];
    }
  }
  scanf("%I64d %I64d", &n, &t);
  while (1ll << (k + 1) <= t) k++;
  if ((1ll << k) != t)
    printf("0");
  else {
    long long ans = 0;
    long long tot = 0, level = 1;
    while (tot + (1ll << level) <= n) {
      ans += cnt[level][k];
      tot += (1ll << level), level++;
    }
    ans += go(n - tot, level, k);
    printf("%I64d", ans);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
long long cnt[100][100];
long long n, t, k;
long long go(long long rest, long long level, long long x) {
  if (level == 1) {
    if (rest == 1)
      return x == 0;
    else
      return x == 1;
  }
  if (rest > (1ll << (level - 1)))
    return go(rest - (1ll << (level - 1)), level - 1, x - 1) +
           cnt[level - 1][x];
  else
    return go(rest, level - 1, x);
}
int main() {
  cnt[1][0] = cnt[1][1] = 1;
  for (int i = 2; i <= 100; ++i) {
    for (int j = 0; j <= i; ++j) {
      cnt[i][j] += cnt[i - 1][j];
      if (j > 0) cnt[i][j] += cnt[i - 1][j - 1];
    }
  }
  scanf("%I64d %I64d", &n, &t);
  while (1ll << (k + 1) <= t) k++;
  if ((1ll << k) != t)
    printf("0");
  else {
    long long ans = 0;
    long long tot = 0, level = 1;
    while (tot + (1ll << level) <= n) {
      ans += cnt[level][k];
      tot += (1ll << level), level++;
    }
    ans += go(n - tot, level, k);
    printf("%I64d", ans);
  }
  return 0;
}

```