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;
struct op {
  long long i, j;
  char c;
  op(long long i, char c, long long j) : i(i), j(j), c(c) {}
  friend ostream& operator<<(ostream& os, const op& a) {
    return os << a.i << " " << a.c << " " << a.j;
  }
};
vector<op> v;
long long mul(long long a, long long b) {
  stack<long long> s;
  long long res = 0;
  for (; b; b >>= 1) {
    if (b & 1) s.push(a);
    v.emplace_back(a, '+', a);
    a += a;
  }
  res = s.top();
  s.pop();
  v.pop_back();
  while (!s.empty()) {
    long long top = s.top();
    s.pop();
    v.emplace_back(res, '+', top);
    res += top;
  }
  return res;
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y;
  y = z - y * (a / b);
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long a;
  cin >> a;
  long long e = log2(a);
  long long b = mul(a, 1 << e);
  v.emplace_back(b, '^', a);
  b ^= a;
  long long x, y;
  long long d = exgcd(a, b, x, y);
  y = -y;
  while (x <= 0 || y <= 0 || y % 2 != 0) {
    x += b;
    y += a;
  }
  long long ax = mul(a, x);
  long long by = mul(b, y);
  v.emplace_back(ax, '^', by);
  assert((a * x) ^ (b * y) == 1);
  cout << v.size() << '\n';
  for (auto& i : v) cout << i << '\n';
  return 0;
}