#include <bits/stdc++.h>
using namespace std;
int cases;
map<long long, int> mp;
int digit[70];
long long dp[70][70];
void init() {
  mp[1] = 1;
  long long t = 1;
  for (int i = 1; i < 44; i++) {
    mp[t << i] = i + 1;
  }
}
long long dfs(int pos, int cnt, bool limit) {
  if (pos == 0) return cnt == 0;
  if (cnt < 0) return 0;
  if (!limit && dp[pos][cnt] != -1) return dp[pos][cnt];
  long long mx = limit ? digit[pos] : 1;
  long long res = 0;
  for (int i = 0; i <= mx; i++) {
    if (i > 1) continue;
    res += dfs(pos - 1, i == 1 ? cnt - 1 : cnt, limit && i == mx);
  }
  return limit ? res : dp[pos][cnt] = res;
}
long long solve(long long n, int k) {
  memset(digit, 0, sizeof(digit));
  int len = 0;
  while (n) {
    digit[++len] = n % 2;
    n /= 2;
  }
  return dfs(len, k, true);
}
int main() {
  long long n, k;
  cin >> n >> k;
  n++;
  init();
  if (mp.find(k) == mp.end()) {
    puts("0");
    return 0;
  }
  memset(dp, -1, sizeof dp);
  long long ans = solve(n, mp[k]);
  if (k == 1) ans--;
  cout << ans << '\n';
}

### 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;
int cases;
map<long long, int> mp;
int digit[70];
long long dp[70][70];
void init() {
  mp[1] = 1;
  long long t = 1;
  for (int i = 1; i < 44; i++) {
    mp[t << i] = i + 1;
  }
}
long long dfs(int pos, int cnt, bool limit) {
  if (pos == 0) return cnt == 0;
  if (cnt < 0) return 0;
  if (!limit && dp[pos][cnt] != -1) return dp[pos][cnt];
  long long mx = limit ? digit[pos] : 1;
  long long res = 0;
  for (int i = 0; i <= mx; i++) {
    if (i > 1) continue;
    res += dfs(pos - 1, i == 1 ? cnt - 1 : cnt, limit && i == mx);
  }
  return limit ? res : dp[pos][cnt] = res;
}
long long solve(long long n, int k) {
  memset(digit, 0, sizeof(digit));
  int len = 0;
  while (n) {
    digit[++len] = n % 2;
    n /= 2;
  }
  return dfs(len, k, true);
}
int main() {
  long long n, k;
  cin >> n >> k;
  n++;
  init();
  if (mp.find(k) == mp.end()) {
    puts("0");
    return 0;
  }
  memset(dp, -1, sizeof dp);
  long long ans = solve(n, mp[k]);
  if (k == 1) ans--;
  cout << ans << '\n';
}

```