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;
struct tri {
  long long a, b, c;
};
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 10;
const long long mod = 1e9 + 7;
const double PI = acos(0) * 2;
int lb(int a) { return a & (-a); }
vector<tri> ans;
long long x, xx, z, y;
void dfs(long long a, long long b) {
  if (a == 1) return;
  while (a != b / 2) {
    ans.push_back({a, a, 1});
    a *= 2;
    if (a & b) {
      ans.push_back({a, b, 2});
      a ^= b;
    }
  }
  if (a & x) {
    ans.push_back({x, a, 2});
    x ^= a;
  }
  dfs(x, a);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> x;
  if (x & 2) {
    ans.push_back({x, x, 1});
    ans.push_back({x, x * 2, 1});
    xx = x * 3;
  } else
    xx = x;
  z = xx;
  while (lb(z * 2) <= xx) {
    ans.push_back({z, z, 1});
    z *= 2;
  }
  ans.push_back({z, xx, 1});
  y = z + xx;
  ans.push_back({y, xx, 2});
  y ^= xx;
  ans.push_back({y, z, 2});
  y ^= z;
  dfs(x, y);
  cout << ans.size() << '\n';
  for (auto i : ans) {
    cout << i.a << " " << (i.c == 1 ? "+" : "^") << " " << i.b << '\n';
  }
  return 0;
}