#include <bits/stdc++.h>
using namespace std;
const int N = 70;
const int MOD = 1000000007;
long long n, t, dp[N][N];
int p = 0;
long long dfs(int l, int c, int limit, long long n) {
  if (l < 0) return c == p;
  if (!limit && dp[l][c] != -1) return dp[l][c];
  int up = limit ? ((n >> l) & 1) : 1;
  long long ret = 0;
  for (int i = 0; i <= up; i++) {
    ret += dfs(l - 1, c + i, limit && i == up, n);
  }
  if (!limit) dp[l][c] = ret;
  return ret;
}
long long gao(long long n) {
  int t = 0;
  while ((1LL << t) < n) t++;
  return dfs(t, 0, 1, n);
}
int main() {
  cin >> n >> t;
  if ((t & (-t)) != t) {
    puts("0");
    return 0;
  }
  while ((1LL << p) < t) p++;
  p++;
  memset(dp, -1, sizeof(dp));
  cout << gao(n + 1) - gao(1) << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 N = 70;
const int MOD = 1000000007;
long long n, t, dp[N][N];
int p = 0;
long long dfs(int l, int c, int limit, long long n) {
  if (l < 0) return c == p;
  if (!limit && dp[l][c] != -1) return dp[l][c];
  int up = limit ? ((n >> l) & 1) : 1;
  long long ret = 0;
  for (int i = 0; i <= up; i++) {
    ret += dfs(l - 1, c + i, limit && i == up, n);
  }
  if (!limit) dp[l][c] = ret;
  return ret;
}
long long gao(long long n) {
  int t = 0;
  while ((1LL << t) < n) t++;
  return dfs(t, 0, 1, n);
}
int main() {
  cin >> n >> t;
  if ((t & (-t)) != t) {
    puts("0");
    return 0;
  }
  while ((1LL << p) < t) p++;
  p++;
  memset(dp, -1, sizeof(dp));
  cout << gao(n + 1) - gao(1) << endl;
  return 0;
}

```