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;
vector<pair<pair<long long, long long>, int> > ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long tmp = n;
  long long pow2n[40];
  pow2n[0] = n;
  for (int i = 0; i < 39; i++) {
    ans.push_back(make_pair(make_pair(tmp, tmp), 0));
    tmp += tmp;
    pow2n[i + 1] = tmp;
  }
  int mxbit = 0;
  for (int i = 0; i < 20; i++) {
    int bit = (1 << i);
    if (n & bit) mxbit = max(mxbit, i);
  }
  ans.push_back(make_pair(make_pair(pow2n[mxbit], n), 1));
  long long coprime = pow2n[mxbit] ^ n;
  tmp = coprime;
  long long pow2coprime[20];
  pow2coprime[0] = coprime;
  for (int i = 0; i < 19; i++) {
    ans.push_back(make_pair(make_pair(tmp, tmp), 0));
    tmp += tmp;
    pow2coprime[i + 1] = tmp;
  }
  long long multCoprime = -1;
  long long multN = -1;
  for (long long i = 1; i <= n; i++) {
    if ((i * coprime) % n == 1) {
      multCoprime = i;
      multN = (i * coprime) / n;
      break;
    }
  }
  assert(multCoprime > 0);
  tmp = -1;
  for (int i = 0; i < 20; i++) {
    int bit = (1 << i);
    if (multCoprime & bit) {
      if (tmp == -1)
        tmp = pow2coprime[i];
      else {
        ans.push_back(make_pair(make_pair(tmp, pow2coprime[i]), 0));
        tmp += pow2coprime[i];
      }
    }
  }
  tmp = -1;
  for (int i = 0; i < 40; i++) {
    long long bit = ((1ll * 1) << (1ll * i));
    if (multN & bit) {
      if (tmp == -1)
        tmp = pow2n[i];
      else {
        ans.push_back(make_pair(make_pair(tmp, pow2n[i]), 0));
        tmp += pow2n[i];
      }
    }
  }
  long long big = coprime * multCoprime;
  long long small = multN * n;
  if (big & 1) {
    ans.push_back(make_pair(make_pair(big, small), 1));
  } else {
    ans.push_back(make_pair(make_pair(big, n), 0));
    ans.push_back(make_pair(make_pair(small, n), 0));
    ans.push_back(make_pair(make_pair(big + n, small + n), 1));
  }
  cout << ans.size() << '\n';
  for (int 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';
    }
  }
}