#include <bits/stdc++.h>
using namespace std;
const int MAXN = 42;
vector<long long> dic[MAXN];
long long ans;
long long N, T;
long long dfs(long long l, long long r, int d, int k) {
  if (k < 0) return 0;
  if (r + 1 == l * 2) return dic[d][k];
  if (l + l / 2 > r) {
    return dfs(l / 2, r - l / 2, d - 1, k);
  } else {
    return dfs(l / 2, l - 1, d - 1, k) + dfs(l / 2, r - l, d - 1, k - 1);
  }
}
int main() {
  dic[0].push_back(1);
  for (int i = 1; i < MAXN; ++i) {
    dic[i].push_back(1);
    dic[0].push_back(0);
  }
  for (int i = 1; i < MAXN; ++i)
    for (int j = 1; j < MAXN; ++j)
      dic[i].push_back(dic[i - 1][j - 1] + dic[i - 1][j]);
  cin >> N >> T;
  ++N;
  int cnt = 0;
  while (T % 2 == 0) {
    T /= 2;
    ++cnt;
  }
  if (T != 1) {
    puts("0");
    return 0;
  }
  T = cnt;
  for (int i = 1;; ++i) {
    if ((1LL << i + 1) > N) {
      ans += dfs(1LL << i, N, i, T);
      break;
    } else {
      ans += dic[i][T];
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAXN = 42;
vector<long long> dic[MAXN];
long long ans;
long long N, T;
long long dfs(long long l, long long r, int d, int k) {
  if (k < 0) return 0;
  if (r + 1 == l * 2) return dic[d][k];
  if (l + l / 2 > r) {
    return dfs(l / 2, r - l / 2, d - 1, k);
  } else {
    return dfs(l / 2, l - 1, d - 1, k) + dfs(l / 2, r - l, d - 1, k - 1);
  }
}
int main() {
  dic[0].push_back(1);
  for (int i = 1; i < MAXN; ++i) {
    dic[i].push_back(1);
    dic[0].push_back(0);
  }
  for (int i = 1; i < MAXN; ++i)
    for (int j = 1; j < MAXN; ++j)
      dic[i].push_back(dic[i - 1][j - 1] + dic[i - 1][j]);
  cin >> N >> T;
  ++N;
  int cnt = 0;
  while (T % 2 == 0) {
    T /= 2;
    ++cnt;
  }
  if (T != 1) {
    puts("0");
    return 0;
  }
  T = cnt;
  for (int i = 1;; ++i) {
    if ((1LL << i + 1) > N) {
      ans += dfs(1LL << i, N, i, T);
      break;
    } else {
      ans += dic[i][T];
    }
  }
  cout << ans << endl;
  return 0;
}

```