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;
long double PI = 3.14159265359;
long long inf = 1000000000000000007;
long long mod = 1000000007;
long long mod1 = 998244353;
const bool multi = 0;
vector<pair<pair<long long, long long>, bool>> ans;
void multiply(long long x, long long y) {
  long long act = x, res = 0;
  while (y > 0) {
    if (y & 1) {
      ans.push_back({{res, act}, 0});
      res += act;
    }
    ans.push_back({{act, act}, 0});
    act += act;
    y /= 2;
  }
}
pair<long long, long long> Euklides(long long a, long long b) {
  if (b == 0) return {1, 0};
  pair<long long, long long> p = Euklides(b, a - (a / b) * b);
  return {p.second, p.first - (a / b) * p.second};
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  if (multi)
    cin >> tt;
  else
    tt = 1;
  while (tt--) {
    long long x, e, y;
    cin >> x;
    ans.push_back({{x, x}, 1});
    for (long long i = 30; i >= 0; i--) {
      if ((x & (1 << i)) > 0) {
        e = (1 << i);
        break;
      }
    }
    y = (e * x) ^ x;
    multiply(x, e);
    ans.push_back({{e * x, x}, 1});
    long long a, b;
    pair<long long, long long> p = Euklides(y, x);
    a = max(abs(p.first), abs(p.second));
    b = min(abs(p.first), abs(p.second));
    while (((a * x) ^ (b * y)) != 1) {
      a += y;
      b += x;
    }
    multiply(x, a);
    multiply(y, b);
    ans.push_back({{x * a, y * b}, 1});
    cout << ans.size() << '\n';
    for (long long i = 0; i < ans.size(); i++) {
      if (ans[i].second)
        cout << ans[i].first.first << " ^ " << ans[i].first.second << '\n';
      else
        cout << ans[i].first.first << " + " << ans[i].first.second << '\n';
    }
  }
  return 0;
}