#include <bits/stdc++.h>
using namespace std;
string tobin(int x, int len) {
  string c;
  while (x > 0) {
    c.push_back(x % 2 + '0');
    x >>= 1;
  }
  while (int((c).size()) < len) c.push_back('0');
  reverse((c).begin(), (c).end());
  return c;
}
long long n, t;
long long dp[45][45];
long long f(int h, int key) {
  if (key < 0 || h < key) return 0;
  if (h == key) return 1;
  long long &ret = dp[h][key];
  if (ret != -1) return ret;
  ret = f(h - 1, key) + f(h - 1, key - 1);
  return ret;
}
long long g(long long idx, int h, int key) {
  if (h < key || key < 0) return 0;
  if (idx + (1LL << h) - 1 <= n) return f(h, key);
  if (idx + (1LL << (h - 1)) - 1 <= n)
    return f(h - 1, key) + g(idx + (1LL << (h - 1)), h - 1, key - 1);
  else
    return g(idx, h - 1, key);
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  if ((t & -t) != t) {
    printf("0\n");
    return 0;
  }
  memset(dp, -1, sizeof(dp));
  int key = 0;
  while (t > 0) {
    key++;
    t >>= 1LL;
  }
  key--;
  long long ans = 0;
  int k = 1;
  while (true) {
    if ((1LL << k) > n) break;
    ans += g(1LL << k, k, key);
    k++;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
string tobin(int x, int len) {
  string c;
  while (x > 0) {
    c.push_back(x % 2 + '0');
    x >>= 1;
  }
  while (int((c).size()) < len) c.push_back('0');
  reverse((c).begin(), (c).end());
  return c;
}
long long n, t;
long long dp[45][45];
long long f(int h, int key) {
  if (key < 0 || h < key) return 0;
  if (h == key) return 1;
  long long &ret = dp[h][key];
  if (ret != -1) return ret;
  ret = f(h - 1, key) + f(h - 1, key - 1);
  return ret;
}
long long g(long long idx, int h, int key) {
  if (h < key || key < 0) return 0;
  if (idx + (1LL << h) - 1 <= n) return f(h, key);
  if (idx + (1LL << (h - 1)) - 1 <= n)
    return f(h - 1, key) + g(idx + (1LL << (h - 1)), h - 1, key - 1);
  else
    return g(idx, h - 1, key);
}
int main() {
  scanf("%I64d%I64d", &n, &t);
  n++;
  if ((t & -t) != t) {
    printf("0\n");
    return 0;
  }
  memset(dp, -1, sizeof(dp));
  int key = 0;
  while (t > 0) {
    key++;
    t >>= 1LL;
  }
  key--;
  long long ans = 0;
  int k = 1;
  while (true) {
    if ((1LL << k) > n) break;
    ans += g(1LL << k, k, key);
    k++;
  }
  printf("%I64d\n", ans);
  return 0;
}

```