#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 60;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
long long n, t;
long long c[maxn][maxn];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t;
  long long ans = 0;
  if (t == 1) ans--;
  if (__builtin_popcountll(t) != 1) cout << 0, exit(0);
  ;
  c[0][0] = 1;
  for (long long i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (long long j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
  t = __builtin_ctzll(t) + 1;
  n += 1;
  long long cnt = 0;
  for (long long i = 50; i >= 0; i--) {
    if (cnt > t) break;
    if (n & (1LL << i)) ans += c[i][t - cnt], cnt++;
  }
  if (__builtin_popcountll(n) == t) ans++;
  cout << ans;
  return (0);
}

### Prompt

Your task is to create a CPP solution 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 60;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
long long n, t;
long long c[maxn][maxn];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t;
  long long ans = 0;
  if (t == 1) ans--;
  if (__builtin_popcountll(t) != 1) cout << 0, exit(0);
  ;
  c[0][0] = 1;
  for (long long i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (long long j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
  t = __builtin_ctzll(t) + 1;
  n += 1;
  long long cnt = 0;
  for (long long i = 50; i >= 0; i--) {
    if (cnt > t) break;
    if (n & (1LL << i)) ans += c[i][t - cnt], cnt++;
  }
  if (__builtin_popcountll(n) == t) ans++;
  cout << ans;
  return (0);
}

```