#include <bits/stdc++.h>
using namespace std;
const long long mod = 10000007;
const long long phi = 9988440;
long long n, a[60], dp[60][60], t;
long long dfs(int id, int pos, int lim, int num, int zero) {
  if (pos == 0) return (!zero) && (num == id);
  if (dp[pos][num] != -1 && !lim) return dp[pos][num];
  int MAX = lim ? a[pos] : 1;
  long long res = 0;
  for (int i = 0; i <= MAX; i++) {
    res += dfs(id, pos - 1, lim && (i == MAX), num + i, zero & (i == 0));
  }
  if (!lim) dp[pos][num] = res;
  return res;
}
long long fans = 0;
long long fp(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    x = x * x;
    y >>= 1;
  }
  return res;
}
void get(long long x) {
  int cnt = 0;
  while (x) {
    a[++cnt] = x & 1;
    x >>= 1;
  }
  memset(dp, -1, sizeof(dp));
  int aim = log2(t) + 1;
  fans = dfs(aim, cnt, 1, 0, 1);
  if (aim == 1) fans--;
  printf("%lld\n", fans);
}
int main() {
  scanf("%lld%lld", &n, &t);
  n++;
  if (fp(2, log2(t)) == t) {
    get(n);
  } else {
    printf("0\n");
    return 0;
  }
  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 mod = 10000007;
const long long phi = 9988440;
long long n, a[60], dp[60][60], t;
long long dfs(int id, int pos, int lim, int num, int zero) {
  if (pos == 0) return (!zero) && (num == id);
  if (dp[pos][num] != -1 && !lim) return dp[pos][num];
  int MAX = lim ? a[pos] : 1;
  long long res = 0;
  for (int i = 0; i <= MAX; i++) {
    res += dfs(id, pos - 1, lim && (i == MAX), num + i, zero & (i == 0));
  }
  if (!lim) dp[pos][num] = res;
  return res;
}
long long fans = 0;
long long fp(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    x = x * x;
    y >>= 1;
  }
  return res;
}
void get(long long x) {
  int cnt = 0;
  while (x) {
    a[++cnt] = x & 1;
    x >>= 1;
  }
  memset(dp, -1, sizeof(dp));
  int aim = log2(t) + 1;
  fans = dfs(aim, cnt, 1, 0, 1);
  if (aim == 1) fans--;
  printf("%lld\n", fans);
}
int main() {
  scanf("%lld%lld", &n, &t);
  n++;
  if (fp(2, log2(t)) == t) {
    get(n);
  } else {
    printf("0\n");
    return 0;
  }
  return 0;
}

```