#include <bits/stdc++.h>
using namespace std;
const int B = 60;
long long n, t;
long long dp[B][B];
int bit[B], b;
long long get(int rem, int d, bool flag) {
  if (rem < 0) return 0;
  if (d == -1) return rem == 0;
  if (!flag && ~dp[rem][d]) return dp[rem][d];
  int lim = flag ? bit[d] : 1;
  long long res = 0;
  for (int i = 0; i <= lim; i++) {
    res += get(rem - i, d - 1, flag && lim == i);
  }
  return flag ? res : dp[rem][d] = res;
}
long long solve(long long x, int cnt) {
  memset(dp, -1, sizeof(dp));
  b = 0;
  long long t = x;
  while (t > 0) {
    bit[b++] = (int)(t % 2);
    t /= 2;
  }
  return get(cnt, b - 1, true);
}
int main() {
  cin >> n >> t;
  if (__builtin_popcountll(t) != 1) return cout << 0 << endl, 0;
  int cnt = 0;
  while (t > 0) {
    cnt++, t /= 2;
  }
  cout << solve(n + 1, cnt) - solve(1, cnt) << endl;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
const int B = 60;
long long n, t;
long long dp[B][B];
int bit[B], b;
long long get(int rem, int d, bool flag) {
  if (rem < 0) return 0;
  if (d == -1) return rem == 0;
  if (!flag && ~dp[rem][d]) return dp[rem][d];
  int lim = flag ? bit[d] : 1;
  long long res = 0;
  for (int i = 0; i <= lim; i++) {
    res += get(rem - i, d - 1, flag && lim == i);
  }
  return flag ? res : dp[rem][d] = res;
}
long long solve(long long x, int cnt) {
  memset(dp, -1, sizeof(dp));
  b = 0;
  long long t = x;
  while (t > 0) {
    bit[b++] = (int)(t % 2);
    t /= 2;
  }
  return get(cnt, b - 1, true);
}
int main() {
  cin >> n >> t;
  if (__builtin_popcountll(t) != 1) return cout << 0 << endl, 0;
  int cnt = 0;
  while (t > 0) {
    cnt++, t /= 2;
  }
  cout << solve(n + 1, cnt) - solve(1, cnt) << endl;
}

```