#include <bits/stdc++.h>
using namespace std;
namespace fastIO1 {
inline char read() {
  static const int IN_LEN = 1000000;
  static char buf[IN_LEN], *s, *t;
  return (s == t               ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin),
          (s == t ? -1 : *s++) : *s++);
}
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = read(); !isdigit(c); c = read()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
inline char readc(char &c) {
  for (c = read(); !isalpha(c) && !isdigit(c); c = read())
    if (c == -1) return 0;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
  if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
  *ooh++ = c;
}
template <class T>
inline void print(T x) {
  static int buf[30], cnt;
  if (x == 0)
    print('0');
  else {
    if (x < 0) print('-'), x = -x;
    for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
    while (cnt) print((char)buf[cnt--]);
  }
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
}  // namespace fastIO1
namespace fastIO2 {
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = getchar(); !isdigit(c); c = getchar()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = getchar()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
}  // namespace fastIO2
using namespace fastIO1;
long long n, ans, f[2][(135)][2];
long long calc(int x, int lena, int lenb, long long sum) {
  memset(f, 0, sizeof(f)), f[1][0][0] = 1;
  int now = 0, pre = 1;
  for (int i = 0; (1ll << i) <= sum; i++, now ^= 1, pre ^= 1) {
    for (int j = 0; j <= x; j++) f[now][j][0] = f[now][j][1] = 0;
    for (int j = 0; j < 2; j++)
      if (i < lena || !j)
        for (int k = 0; k < 2; k++)
          if (i < lenb || !k)
            for (int l = 0; l < 2; l++) {
              int S = j + k + l;
              if (((sum >> i) & 1) ^ (S & 1)) continue;
              for (int m = 0; m + j + k <= x; m++)
                f[now][m + j + k][S >> 1] += f[pre][m][l];
            }
  }
  return f[pre][x][0];
}
int main() {
  read(n);
  for (int i = 0; i <= 60; i++)
    for (int j = 0; j <= 60; j++) {
      long long mul = (1ll << i + 1) + (1ll << j + 1) - 3;
      if (n < mul) break;
      long long ret = n % mul - (1ll << j) + 1;
      if (ret < 0) continue;
      for (int k = (ret & 1); k <= i + j; k += 2)
        ans += calc(k, i - 1, j - 1, (ret + k) >> 1);
    }
  cout << ans;
}

### Prompt

Create a solution in cpp for the following problem:

The New Year tree is an infinite perfect binary tree rooted in the node 1. Each node v has two children: nodes indexed (2·v) and (2·v + 1).

<image>

Polar bears love decorating the New Year tree and Limak is no exception. As he is only a little bear, he was told to decorate only one simple path between some pair of nodes. Though he was given an opportunity to pick the pair himself! Now he wants to know the number of unordered pairs of indices (u, v) (u ≤ v), such that the sum of indices of all nodes along the simple path between u and v (including endpoints) is equal to s. Can you help him and count this value?

Input

The only line of the input contains a single integer s (1 ≤ s ≤ 1015).

Output

Print one integer, denoting the number of unordered pairs of nodes indices defining simple paths with the sum of indices of vertices equal to s.

Example

Input

10


Output

4

Note

In sample test, there are 4 paths with the sum of indices equal to 10:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
namespace fastIO1 {
inline char read() {
  static const int IN_LEN = 1000000;
  static char buf[IN_LEN], *s, *t;
  return (s == t               ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin),
          (s == t ? -1 : *s++) : *s++);
}
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = read(); !isdigit(c); c = read()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
inline char readc(char &c) {
  for (c = read(); !isalpha(c) && !isdigit(c); c = read())
    if (c == -1) return 0;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
  if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
  *ooh++ = c;
}
template <class T>
inline void print(T x) {
  static int buf[30], cnt;
  if (x == 0)
    print('0');
  else {
    if (x < 0) print('-'), x = -x;
    for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
    while (cnt) print((char)buf[cnt--]);
  }
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
}  // namespace fastIO1
namespace fastIO2 {
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = getchar(); !isdigit(c); c = getchar()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = getchar()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
}  // namespace fastIO2
using namespace fastIO1;
long long n, ans, f[2][(135)][2];
long long calc(int x, int lena, int lenb, long long sum) {
  memset(f, 0, sizeof(f)), f[1][0][0] = 1;
  int now = 0, pre = 1;
  for (int i = 0; (1ll << i) <= sum; i++, now ^= 1, pre ^= 1) {
    for (int j = 0; j <= x; j++) f[now][j][0] = f[now][j][1] = 0;
    for (int j = 0; j < 2; j++)
      if (i < lena || !j)
        for (int k = 0; k < 2; k++)
          if (i < lenb || !k)
            for (int l = 0; l < 2; l++) {
              int S = j + k + l;
              if (((sum >> i) & 1) ^ (S & 1)) continue;
              for (int m = 0; m + j + k <= x; m++)
                f[now][m + j + k][S >> 1] += f[pre][m][l];
            }
  }
  return f[pre][x][0];
}
int main() {
  read(n);
  for (int i = 0; i <= 60; i++)
    for (int j = 0; j <= 60; j++) {
      long long mul = (1ll << i + 1) + (1ll << j + 1) - 3;
      if (n < mul) break;
      long long ret = n % mul - (1ll << j) + 1;
      if (ret < 0) continue;
      for (int k = (ret & 1); k <= i + j; k += 2)
        ans += calc(k, i - 1, j - 1, (ret + k) >> 1);
    }
  cout << ans;
}

```