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;
using ll = long long;
using cd = complex<double>;
vector<string> v;
ll popcount(ll x) {
  ll i = 0;
  for (; x >= (1LL << i); i++)
    ;
  return --i;
}
pair<ll, ll> gcd_extended(ll a, ll b) {
  ll a1 = 1, b1 = 0, a2 = 0, b2 = (a / b);
  ll r = a % b;
  while (r > 1) {
    ll q = (b / r);
    ll t1 = (a1 - a2) * q, t2 = (b1 - b2) * q;
    a1 = a2 / (a / b), b1 = b2 / (a / b);
    a2 = t1, b2 = t2;
    a = b;
    b = r;
    r = a % b;
  }
  assert(r == 1);
  return make_pair(a1 - a2, b1 - b2);
}
void mult(ll a, ll b) {
  ll st = 0;
  while (b) {
    if (b & 1) {
      if (st && a) {
        v.push_back(to_string(st).append(" + ").append(to_string(a)));
      }
      st += a;
    }
    if (a) {
      v.push_back(to_string(a).append(" + ").append(to_string(a)));
    }
    a += a;
    b >>= 1;
  }
}
int main() {
  ll n;
  cin >> n;
  mult(n, (1LL << popcount(n)));
  ll y = (n << popcount(n)) ^ n;
  v.push_back(to_string((n << popcount(n))).append(" ^ ").append(to_string(n)));
  pair<ll, ll> p = gcd_extended(y, n);
  if (p.first < 0) {
    swap(y, n);
    swap(p.first, p.second);
  }
  p.second *= -1;
  if (p.second & 1) {
    p.first += n, p.second += y;
  }
  mult(y, p.first);
  mult(n, p.second);
  v.push_back(
      to_string(y * p.first).append(" ^ ").append(to_string(n * p.second)));
  cout << v.size() << "\n";
  for (string s : v) {
    cout << s << "\n";
  }
}