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;
const int inf = 1e9 + 7;
const long long INF = 1e18;
vector<pair<int, pair<long long, long long> > > ov;
void make(long long x, long long mul) {
  if (mul == 1) return;
  if (mul % 2 == 0) {
    make(x, mul / 2);
    ov.emplace_back(1, pair<long long, long long>(x * mul / 2, x * mul / 2));
  } else {
    make(x, (mul - 1) / 2);
    ov.emplace_back(
        1, pair<long long, long long>(x * (mul - 1) / 2, x * (mul - 1) / 2));
    ov.emplace_back(1, pair<long long, long long>(x * (mul - 1), x));
  }
  return;
}
inline long long reduce(long long x) {
  int p = 0;
  while (x > (1ll << p)) p++;
  long long m1 = 1ll << p;
  long long m2 = m1 - 1;
  make(x, m1);
  make(x, m2);
  ov.emplace_back(2, pair<long long, long long>(x * m1, x * m2));
  ov.emplace_back(1, pair<long long, long long>(m1 + m1 - x, x));
  long long y = x;
  while (y != m1) {
    ov.emplace_back(1, pair<long long, long long>(y, y));
    y *= 2;
    if (y >= m1 + m1) {
      ov.emplace_back(2, pair<long long, long long>(y, m1 + m1));
      y -= m1 + m1;
    }
  }
  ov.emplace_back(2, pair<long long, long long>(m1 + m1 - x, m1));
  return m1 - x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long x;
  cin >> x;
  while (x != 1) x = reduce(x);
  cout << ov.size() << endl;
  for (auto& t : ov) {
    if (t.first == 1)
      cout << t.second.first << " + " << t.second.second << '\n';
    else
      cout << t.second.first << " ^ " << t.second.second << '\n';
  }
  return 0;
}