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;
constexpr long long int MOD = 1000000007;
constexpr double EPS = 1e-12;
long long int N, M, K, T, H, W, L, R;
vector<string> ans;
void add(long long int a, long long int b) {
  ans.push_back(to_string(a) + " + " + to_string(b));
}
void XOR(long long int a, long long int b) {
  ans.push_back(to_string(a) + " ^ " + to_string(b));
}
int func(int num) {
  int ret = num;
  vector<long long int> by(21, num);
  for (int i = 1; i <= 20; i++) {
    by[i] = by[i - 1] * 2;
    add(by[i - 1], by[i - 1]);
  }
  int idx = 0;
  for (int i = 1; i <= 20; i++) {
    if (by[i] & by[0]) {
      idx = i;
      break;
    }
  }
  int tp = 0;
  for (int i = 0; i < 20; i++) {
    if ((num >> i) & 1) {
      tp = i;
    }
  }
  long long int a = by[idx];
  long long int b = by[idx] ^ by[0];
  XOR(by[idx], by[0]);
  for (int i = 1; i < 1 << 20; i++) {
    if ((min(a, b) + num * i) % (1 << (tp + 1)) == 0) {
      for (int j = 0; j < 20; j++) {
        if ((i >> j) & 1) {
          add(a, by[j]);
          add(b, by[j]);
          a += by[j];
          b += by[j];
        }
      }
      ret = a ^ b;
      XOR(a, b);
      break;
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  while (N > 1) {
    N = func(N);
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i << endl;
}