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;
unordered_map<long long, long long> num;
vector<long long> v;
vector<pair<long long, pair<long long, long long> > > res;
void go2(long long x, long long k) {
  for (long long i = 0; i < 21; ++i) {
    res.push_back({0, {(1 << i) * x, (1 << i) * x}});
  }
  vector<long long> v;
  for (long long i = 0; i < 21; ++i) {
    if ((1 << i) & k) {
      v.push_back(i);
    }
  }
  long long curr = 0;
  for (long long i = 0; i < v.size(); ++i) {
    res.push_back({0, {curr, (1 << v[i]) * x}});
    curr += ((1 << v[i]) * x);
  }
}
void go1(long long x) {
  for (long long k = 0; k < (1 << 20); ++k) {
    if (((x * k) % (1 << 20)) == 1) {
      go2(x, k);
      go2(x, k + 1);
      res.push_back({1, {x * k, x * (k + 1)}});
      break;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  res.push_back({1, {x, x}});
  go1(x);
  go1(x + 2);
  if ((x ^ (x + 2)) == 2) {
    res.push_back({1, {x, x + 2}});
  } else {
    res.push_back({1, {x + 2, x + 4}});
  }
  go2(2, (x - 1) / 2);
  res.push_back({1, {x - 1, x}});
  cout << res.size() << endl;
  for (auto h : res) {
    if (h.first == 0) {
      cout << h.second.first << ' ' << "+" << ' ' << h.second.second << endl;
    } else {
      cout << h.second.first << ' ' << "^" << ' ' << h.second.second << endl;
    }
  }
  return 0;
}