#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;
long long a[67];
long long dp[67][67];
long long n, ju, t, res, base;
long long Count(long long k, long long tmp) {
  if (k < tmp) return dp[k][tmp] = 0;
  if (!tmp || k == tmp) {
    return dp[k][tmp] = 1;
  }
  if (dp[k][tmp] != -1) return dp[k][tmp];
  return dp[k][tmp] = Count(k - 1, tmp - 1) + Count(k - 1, tmp);
}
void init() {
  ju = n;
  for (long long i = 0, j = 1; j <= ju; j <<= 1, i++) {
    res += Count(i, base);
    ju -= j;
  }
  long long tp = 0;
  for (long long i = 63; i >= 0; i--) {
    if ((ju >> i) & 1) {
      res += Count(i, base - tp);
      tp++;
      if (tp > base) break;
    }
  }
}
int main() {
  memset(dp, -1, sizeof dp);
  while (cin >> n >> t) {
    memset(a, 0, sizeof a);
    n++;
    res = 0;
    base = 0;
    while (t > 1) {
      base++;
      if (t & 1) {
        puts("0");
        return 0;
      }
      t /= 2;
    }
    init();
    if (base == 0) res--;
    cout << res << endl;
  }
  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;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;
long long a[67];
long long dp[67][67];
long long n, ju, t, res, base;
long long Count(long long k, long long tmp) {
  if (k < tmp) return dp[k][tmp] = 0;
  if (!tmp || k == tmp) {
    return dp[k][tmp] = 1;
  }
  if (dp[k][tmp] != -1) return dp[k][tmp];
  return dp[k][tmp] = Count(k - 1, tmp - 1) + Count(k - 1, tmp);
}
void init() {
  ju = n;
  for (long long i = 0, j = 1; j <= ju; j <<= 1, i++) {
    res += Count(i, base);
    ju -= j;
  }
  long long tp = 0;
  for (long long i = 63; i >= 0; i--) {
    if ((ju >> i) & 1) {
      res += Count(i, base - tp);
      tp++;
      if (tp > base) break;
    }
  }
}
int main() {
  memset(dp, -1, sizeof dp);
  while (cin >> n >> t) {
    memset(a, 0, sizeof a);
    n++;
    res = 0;
    base = 0;
    while (t > 1) {
      base++;
      if (t & 1) {
        puts("0");
        return 0;
      }
      t /= 2;
    }
    init();
    if (base == 0) res--;
    cout << res << endl;
  }
  return 0;
}

```