Problem: You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.

You may write new numbers on the blackboard with the following two operations. 

  * You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard. 
  * You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard. 

Perform a sequence of operations such that at the end the number 1 is on the blackboard.

Input

The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).

Output

Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation. 

  * The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard. 
  * The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard. 

The operation symbol (+ or ^) must be separated from a, b by a whitespace.

You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.

Examples

Input


3


Output


5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9


Input


123


Output


10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  static char ch;
  bool f = 1;
  for (x = 0, ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = 0;
  for (; isdigit(ch); x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar())
    ;
  x = f ? x : -x;
}
template <typename T>
void print(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void print(T x, char ap) {
  print(x);
  if (ap) putchar(ap);
}
template <typename T>
inline void chkmax(T &x, const T &y) {
  x = x < y ? y : x;
}
template <typename T>
inline void chkmin(T &x, const T &y) {
  x = x < y ? x : y;
}
vector<pair<pair<long long, int>, long long> > v;
int n;
inline int doit(int n) {
  long long now = n;
  int ans = 0;
  while (now > 1) ++ans, now >>= 1;
  now = n;
  for (register int i = 1, _r = ans; i <= _r; ++i)
    v.push_back(make_pair(make_pair(now, 0), now)), now <<= 1;
  long long z = n ^ now;
  v.push_back(make_pair(make_pair(n, 1), now));
  long long r = now + z;
  v.push_back(make_pair(make_pair(now, 0), z));
  v.push_back(make_pair(make_pair(now, 0), now));
  long long t = (now << 1) ^ r;
  v.push_back(make_pair(make_pair(now << 1, 1), r));
  v.push_back(make_pair(make_pair(t, 1), n)), t ^= n;
  long long s = t;
  while (now > s) {
    if (now & t) v.push_back(make_pair(make_pair(now, 1), t)), now ^= t;
    v.push_back(make_pair(make_pair(t, 0), t)), t <<= 1;
  }
  v.push_back(make_pair(make_pair(now, 1), n)), n ^= now;
  return n;
}
int main() {
  read(n);
  while (n > 1) n = doit(n);
  print(v.size(), '\n');
  for (auto &p : v)
    print(p.first.first, ' '), putchar(p.first.second ? '^' : '+'),
        putchar(' '), print(p.second, '\n');
  return 0;
}