#include <bits/stdc++.h>
long long n, t;
long long memo[66][66][2];
bool done[66][66][2];
int countBits(long long x) {
  int res = 0;
  while (x) {
    res++;
    x >>= 1;
  }
  return res;
}
long long dp(int i, int j, int s) {
  if (done[i][j][s]) return memo[i][j][s];
  done[i][j][s] = true;
  long long &ret = memo[i][j][s] = 0;
  if (i == 0) return ret = (j == 0);
  for (int x = 0, bound = s ? (((n) >> (i - 1)) & 1) : 1; x <= bound; x++) {
    if (j - (x == 1) >= 0) ret += dp(i - 1, j - (x == 1), s && (x == bound));
  }
  return ret;
}
int main() {
  scanf(
      "%lld"
      "%lld",
      &n, &t);
  n++;
  int lg = countBits(t);
  if ((1LL << (lg - 1)) != t) {
    puts("0");
    return 0;
  }
  printf(
      "%lld"
      "\n",
      dp(countBits(n), lg, 1) - (t == 1 ? 1 : 0));
}

### Prompt

Your challenge is to write 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>
long long n, t;
long long memo[66][66][2];
bool done[66][66][2];
int countBits(long long x) {
  int res = 0;
  while (x) {
    res++;
    x >>= 1;
  }
  return res;
}
long long dp(int i, int j, int s) {
  if (done[i][j][s]) return memo[i][j][s];
  done[i][j][s] = true;
  long long &ret = memo[i][j][s] = 0;
  if (i == 0) return ret = (j == 0);
  for (int x = 0, bound = s ? (((n) >> (i - 1)) & 1) : 1; x <= bound; x++) {
    if (j - (x == 1) >= 0) ret += dp(i - 1, j - (x == 1), s && (x == bound));
  }
  return ret;
}
int main() {
  scanf(
      "%lld"
      "%lld",
      &n, &t);
  n++;
  int lg = countBits(t);
  if ((1LL << (lg - 1)) != t) {
    puts("0");
    return 0;
  }
  printf(
      "%lld"
      "\n",
      dp(countBits(n), lg, 1) - (t == 1 ? 1 : 0));
}

```