#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
map<long long, map<long long, long long> > Map;
inline int log2(long long n) {
  int cnt = 0;
  while (n) n >>= 1, cnt++;
  return cnt - 1;
}
long long find(long long n, long long t) {
  if (t == 1) {
    int i;
    for (i = 0; (1LL << i) <= n; i++)
      ;
    return i;
  }
  if (t >= (n + 1)) return 0;
  if (Map[n][t] != 0) return Map[n][t];
  int c = (log2(n));
  long long s = (1LL << c - 1) - 1;
  long long ans = 0;
  ans += find(n - (1LL << c), t / 2);
  ans += find(s, t) + find(s, t / 2);
  return Map[n][t] = ans;
}
int main() {
  long long n, t;
  while (~scanf("%I64d%I64d", &n, &t)) {
    if ((1LL << log2(t)) != t) {
      puts("0");
      continue;
    }
    n++;
    if (t == 1) {
      int i;
      for (i = 0; (1LL << i) <= n; i++)
        ;
      printf("%d\n", i - 1);
    } else {
      long long ans = find(n, t);
      printf("%I64d\n", ans);
    }
  }
  scanf(" ");
}

### Prompt

Generate 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>
using namespace std;
const int N = 100010;
map<long long, map<long long, long long> > Map;
inline int log2(long long n) {
  int cnt = 0;
  while (n) n >>= 1, cnt++;
  return cnt - 1;
}
long long find(long long n, long long t) {
  if (t == 1) {
    int i;
    for (i = 0; (1LL << i) <= n; i++)
      ;
    return i;
  }
  if (t >= (n + 1)) return 0;
  if (Map[n][t] != 0) return Map[n][t];
  int c = (log2(n));
  long long s = (1LL << c - 1) - 1;
  long long ans = 0;
  ans += find(n - (1LL << c), t / 2);
  ans += find(s, t) + find(s, t / 2);
  return Map[n][t] = ans;
}
int main() {
  long long n, t;
  while (~scanf("%I64d%I64d", &n, &t)) {
    if ((1LL << log2(t)) != t) {
      puts("0");
      continue;
    }
    n++;
    if (t == 1) {
      int i;
      for (i = 0; (1LL << i) <= n; i++)
        ;
      printf("%d\n", i - 1);
    } else {
      long long ans = find(n, t);
      printf("%I64d\n", ans);
    }
  }
  scanf(" ");
}

```