#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long c[65][65], ans;
int cnt;
struct Tprogram {
  void open() {
    freopen("", "r", stdin);
    freopen("", "w", stdout);
  }
  void close() {
    fclose(stdin);
    fclose(stdout);
  }
  void init() {
    for (int i = 0; i <= 60; i++) c[i][0] = 1;
    for (int i = 1; i <= 60; i++)
      for (int j = 1; j <= 60; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    scanf("%I64d%I64d", &n, &t);
    n++;
  }
  long long ask(long long k, long long n, int C) {
    if (n == 0) return 0;
    if (C == 0) return 1;
    if (k == 0) return 0;
    long long s = 1LL << k - 1;
    if (n >= s)
      return c[k - 1][C] + ask(k - 1, n - s, C - 1);
    else
      return ask(k - 1, n, C);
  }
  void work() {
    if ((t & -t) != t)
      printf("0\n");
    else {
      cnt = 0;
      while (t != 1) cnt++, t >>= 1;
      for (int i = 0; i <= 60; i++) {
        if ((1LL << i) <= n)
          ans += c[i][cnt];
        else {
          ans += ask(i, n, cnt);
          break;
        }
        n -= (1LL << i);
      }
      if (cnt == 0) ans--;
      printf("%I64d\n", ans);
    }
  }
} Program;
int main() {
  Program.init();
  Program.work();
  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>
using namespace std;
long long n, t;
long long c[65][65], ans;
int cnt;
struct Tprogram {
  void open() {
    freopen("", "r", stdin);
    freopen("", "w", stdout);
  }
  void close() {
    fclose(stdin);
    fclose(stdout);
  }
  void init() {
    for (int i = 0; i <= 60; i++) c[i][0] = 1;
    for (int i = 1; i <= 60; i++)
      for (int j = 1; j <= 60; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    scanf("%I64d%I64d", &n, &t);
    n++;
  }
  long long ask(long long k, long long n, int C) {
    if (n == 0) return 0;
    if (C == 0) return 1;
    if (k == 0) return 0;
    long long s = 1LL << k - 1;
    if (n >= s)
      return c[k - 1][C] + ask(k - 1, n - s, C - 1);
    else
      return ask(k - 1, n, C);
  }
  void work() {
    if ((t & -t) != t)
      printf("0\n");
    else {
      cnt = 0;
      while (t != 1) cnt++, t >>= 1;
      for (int i = 0; i <= 60; i++) {
        if ((1LL << i) <= n)
          ans += c[i][cnt];
        else {
          ans += ask(i, n, cnt);
          break;
        }
        n -= (1LL << i);
      }
      if (cnt == 0) ans--;
      printf("%I64d\n", ans);
    }
  }
} Program;
int main() {
  Program.init();
  Program.work();
  return 0;
}

```