#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
map<pair<long long, long long>, long long> d;
long long solve(long long n, long long t) {
  if (d.count(make_pair(n, t))) return d[make_pair(n, t)];
  if (n == 0) return d[make_pair(n, t)] = 0;
  long long l = 1, p = 0;
  while (2 * l <= n) {
    l *= 2;
    p++;
  }
  if (t == 1) return d[make_pair(n, t)] = p + 1;
  if (n <= 2) return d[make_pair(n, t)] = 0;
  long long ans = solve(l / 2 - 1, t);
  if (t % 2 == 0) ans += solve(l / 2 - 1, t / 2);
  if (n != l && t % 2 == 0) ans += solve(n - l, t / 2);
  return d[make_pair(n, t)] = ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  long long ans = solve(n + 1, t);
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
template <typename T>
inline T Abs(T x) {
  return (x >= 0) ? x : -x;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
string toStr(T x) {
  stringstream st;
  st << x;
  string s;
  st >> s;
  return s;
}
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
map<pair<long long, long long>, long long> d;
long long solve(long long n, long long t) {
  if (d.count(make_pair(n, t))) return d[make_pair(n, t)];
  if (n == 0) return d[make_pair(n, t)] = 0;
  long long l = 1, p = 0;
  while (2 * l <= n) {
    l *= 2;
    p++;
  }
  if (t == 1) return d[make_pair(n, t)] = p + 1;
  if (n <= 2) return d[make_pair(n, t)] = 0;
  long long ans = solve(l / 2 - 1, t);
  if (t % 2 == 0) ans += solve(l / 2 - 1, t / 2);
  if (n != l && t % 2 == 0) ans += solve(n - l, t / 2);
  return d[make_pair(n, t)] = ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  long long ans = solve(n + 1, t);
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

```