#include <bits/stdc++.h>
using namespace std;
const long long maxn = 55;
const long long mod = 10000007;
long long n, t, len, a[maxn], f[maxn][maxn];
void get(long long x, bool pa) {
  len = 0;
  while (x) {
    if (pa) a[len + 1] = x % 2;
    x >>= 1;
    len++;
  }
}
long long dp(long long pos, long long num, long long ask, bool limit) {
  if (pos == 0) return num == ask;
  if (f[pos][num] != -1 && !limit) return f[pos][num];
  long long up = limit ? a[pos] : 1;
  long long ans = 0;
  for (long long i = 0; i <= up; i++) {
    ans += dp(pos - 1, num + i, ask, (limit && (i == up)));
  }
  return limit ? ans : f[pos][num] = ans;
}
long long fsm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res % mod;
}
int main() {
  cin >> n >> t;
  n++;
  get(n, 1);
  if (t - (t & (-t)) != 0) {
    puts("0");
    return 0;
  }
  memset(f, -1, sizeof f);
  long long ans = dp(len, 0, log2(t) + 1, 1);
  if (t == 1) ans--;
  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 long long maxn = 55;
const long long mod = 10000007;
long long n, t, len, a[maxn], f[maxn][maxn];
void get(long long x, bool pa) {
  len = 0;
  while (x) {
    if (pa) a[len + 1] = x % 2;
    x >>= 1;
    len++;
  }
}
long long dp(long long pos, long long num, long long ask, bool limit) {
  if (pos == 0) return num == ask;
  if (f[pos][num] != -1 && !limit) return f[pos][num];
  long long up = limit ? a[pos] : 1;
  long long ans = 0;
  for (long long i = 0; i <= up; i++) {
    ans += dp(pos - 1, num + i, ask, (limit && (i == up)));
  }
  return limit ? ans : f[pos][num] = ans;
}
long long fsm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res % mod;
}
int main() {
  cin >> n >> t;
  n++;
  get(n, 1);
  if (t - (t & (-t)) != 0) {
    puts("0");
    return 0;
  }
  memset(f, -1, sizeof f);
  long long ans = dp(len, 0, log2(t) + 1, 1);
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

```