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>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int P = 20;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int64_t X;
  cin >> X;
  vector<string> ops;
  ops.push_back(to_string(X) + " ^ " + to_string(X));
  while (X != 1) {
    vector<int64_t> numbers = {0, X};
    for (int p = 0; p < P; p++) {
      ops.push_back(to_string(X << p) + " + " + to_string(X << p));
      numbers.push_back(X << (p + 1));
    }
    for (int p = 1; p < P; p++) {
      ops.push_back(to_string(X * ((1 << p) - 1)) + " + " + to_string(X << p));
      numbers.push_back(X * ((1 << (p + 1)) - 1));
    }
    sort(numbers.begin(), numbers.end());
    int n = int(numbers.size());
    int64_t best = X;
    vector<int64_t> choices;
    for (int i = 0; i < n; i++)
      for (int j = i + 1; j < n; j++)
        for (int k = j + 1; k < n; k++) {
          int64_t candidate = numbers[i] ^ numbers[j] ^ numbers[k];
          if (candidate % 2 != 0 && candidate < best) {
            best = candidate;
            choices = {numbers[i], numbers[j], numbers[k]};
          }
        }
    assert(best < X);
    ops.push_back(to_string(choices[0]) + " ^ " + to_string(choices[1]));
    ops.push_back(to_string(choices[0] ^ choices[1]) + " ^ " +
                  to_string(choices[2]));
    X = best;
  }
  cout << ops.size() << '\n';
  for (auto &op : ops) cout << op << '\n';
}