#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-7;
const int inf = (1 << 31) - 1;
const int hinf = 0x3f3f3f3f;
const long long mod = 1000000007;
long long n, k;
int pos;
int ff;
long long two[64];
long long dp[64][64][3];
int bit[64];
int cnt;
void ini() {
  two[0] = 1ll;
  for (int i = 1; i < 63; i++) two[i] = two[i - 1] * 2ll;
  ff = 0;
  for (int i = 0; i < 63; i++)
    if (two[i] == k) {
      ff = 1;
      pos = i;
    }
}
int cc(long long x) { return x == 0 ? 0 : cc(x >> 1) + 1; }
int main() {
  cin >> n >> k;
  n++;
  ini();
  if (ff == 0) {
    puts("0");
    return 0;
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1ll;
  cnt = cc(n);
  long long nn = n;
  for (int i = 1; i <= cnt; i++) {
    bit[i] = nn & 1;
    nn >>= 1;
  }
  reverse(bit + 1, bit + 1 + cnt);
  for (int i = 1; i <= cnt; i++) {
    if (i == 1) {
      dp[1][1][1] = 1ll;
    } else {
      for (int j = 1; j < i; j++) {
        for (int k = 0; k < 3; k++) {
          if (dp[i - 1][j][k]) {
            if (k == 1) {
              if (bit[i]) {
                dp[i][j + 1][k] += dp[i - 1][j][k];
                dp[i][j][0] += dp[i - 1][j][k];
              } else {
                dp[i][j][k] += dp[i - 1][j][k];
                dp[i][j + 1][2] += dp[i - 1][j][k];
              }
            } else if (k == 0) {
              dp[i][j + 1][0] += dp[i - 1][j][k];
              dp[i][j][0] += dp[i - 1][j][k];
            } else if (k == 2) {
              dp[i][j + 1][2] += dp[i - 1][j][k];
              dp[i][j][2] += dp[i - 1][j][k];
            }
          }
        }
      }
    }
  }
  long long ans = 0;
  if (pos == 0) ans--;
  for (int i = pos + 1; i <= cnt; i++) {
    if (i == cnt)
      for (int k = 0; k < 2; k++) ans += dp[i][pos + 1][k];
    else {
      for (int k = 0; k < 3; k++) ans += dp[i][pos + 1][k];
    }
  }
  cout << ans << endl;
}

### Prompt

Generate 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;
const double eps = 1e-7;
const int inf = (1 << 31) - 1;
const int hinf = 0x3f3f3f3f;
const long long mod = 1000000007;
long long n, k;
int pos;
int ff;
long long two[64];
long long dp[64][64][3];
int bit[64];
int cnt;
void ini() {
  two[0] = 1ll;
  for (int i = 1; i < 63; i++) two[i] = two[i - 1] * 2ll;
  ff = 0;
  for (int i = 0; i < 63; i++)
    if (two[i] == k) {
      ff = 1;
      pos = i;
    }
}
int cc(long long x) { return x == 0 ? 0 : cc(x >> 1) + 1; }
int main() {
  cin >> n >> k;
  n++;
  ini();
  if (ff == 0) {
    puts("0");
    return 0;
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1ll;
  cnt = cc(n);
  long long nn = n;
  for (int i = 1; i <= cnt; i++) {
    bit[i] = nn & 1;
    nn >>= 1;
  }
  reverse(bit + 1, bit + 1 + cnt);
  for (int i = 1; i <= cnt; i++) {
    if (i == 1) {
      dp[1][1][1] = 1ll;
    } else {
      for (int j = 1; j < i; j++) {
        for (int k = 0; k < 3; k++) {
          if (dp[i - 1][j][k]) {
            if (k == 1) {
              if (bit[i]) {
                dp[i][j + 1][k] += dp[i - 1][j][k];
                dp[i][j][0] += dp[i - 1][j][k];
              } else {
                dp[i][j][k] += dp[i - 1][j][k];
                dp[i][j + 1][2] += dp[i - 1][j][k];
              }
            } else if (k == 0) {
              dp[i][j + 1][0] += dp[i - 1][j][k];
              dp[i][j][0] += dp[i - 1][j][k];
            } else if (k == 2) {
              dp[i][j + 1][2] += dp[i - 1][j][k];
              dp[i][j][2] += dp[i - 1][j][k];
            }
          }
        }
      }
    }
  }
  long long ans = 0;
  if (pos == 0) ans--;
  for (int i = pos + 1; i <= cnt; i++) {
    if (i == cnt)
      for (int k = 0; k < 2; k++) ans += dp[i][pos + 1][k];
    else {
      for (int k = 0; k < 3; k++) ans += dp[i][pos + 1][k];
    }
  }
  cout << ans << endl;
}

```