#include <bits/stdc++.h>
using namespace std;
const long long MAX_N = 10005;
const long long MAX_D = 105;
string k;
long long D;
long long dp[MAX_N][MAX_D];
long long dfs(long long pos, long long sum, bool limit) {
  if (pos == k.size()) return sum != D ? 0 : 1;
  if (!limit && dp[pos][sum] != -1) return dp[pos][sum];
  long long u = limit ? k[pos] - '0' : 1;
  long long ret = 0;
  for (long long i = 0; i <= u; ++i) {
    ret += dfs(pos + 1, sum + i, limit && i == k[pos] - '0');
  }
  if (!limit) dp[pos][sum] = ret;
  return ret;
}
signed main() {
  memset(dp, -1, sizeof(dp));
  long long n, t;
  cin >> n >> t;
  if (t != ((t) & (-(t)))) {
    puts("0");
    return 0;
  }
  D = __builtin_ffsll(t);
  n++;
  for (long long i = 63; i >= 0; i--) {
    if (n & (1ll << i))
      k.push_back('1');
    else
      k.push_back('0');
  }
  long long ans = dfs(0, 0, true);
  if (D == 1) ans--;
  cout << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 long long MAX_N = 10005;
const long long MAX_D = 105;
string k;
long long D;
long long dp[MAX_N][MAX_D];
long long dfs(long long pos, long long sum, bool limit) {
  if (pos == k.size()) return sum != D ? 0 : 1;
  if (!limit && dp[pos][sum] != -1) return dp[pos][sum];
  long long u = limit ? k[pos] - '0' : 1;
  long long ret = 0;
  for (long long i = 0; i <= u; ++i) {
    ret += dfs(pos + 1, sum + i, limit && i == k[pos] - '0');
  }
  if (!limit) dp[pos][sum] = ret;
  return ret;
}
signed main() {
  memset(dp, -1, sizeof(dp));
  long long n, t;
  cin >> n >> t;
  if (t != ((t) & (-(t)))) {
    puts("0");
    return 0;
  }
  D = __builtin_ffsll(t);
  n++;
  for (long long i = 63; i >= 0; i--) {
    if (n & (1ll << i))
      k.push_back('1');
    else
      k.push_back('0');
  }
  long long ans = dfs(0, 0, true);
  if (D == 1) ans--;
  cout << ans << endl;
  return 0;
}

```