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>
#pragma comment(linker, "/STACK:1000000000")
#pragma GCC optimize("Ofast")
using namespace std;
const int maxlog = 61;
vector<pair<int, long long> > cur;
vector<pair<long long, long long> > ans;
bool add(long long x, bool nd = true) {
  for (auto [b, y] : cur) {
    if ((x >> b) & 1) {
      ans.push_back(make_pair(x, y));
      x ^= y;
    }
  }
  if (x == 0) {
    return false;
  }
  if (nd) {
    int b = maxlog - 1;
    while (!((x >> b) & 1)) {
      b--;
    }
    for (auto &[f, y] : cur) {
      if ((y >> b) & 1) {
        ans.push_back(make_pair(y, x));
        y ^= x;
      }
    }
    cur.push_back(make_pair(b, x));
  }
  return true;
}
vector<pair<char, pair<long long, long long> > > res;
int main() {
  int x;
  cin >> x;
  long long z = x;
  for (int i = 0; i < 30; i++) {
    res.push_back(make_pair('+', make_pair(z, z)));
    z *= 2;
  }
  for (int i = 1; i <= (int)1e7; i++) {
    ans.clear();
    if (add((long long)i * x)) {
      long long cur = 0;
      for (int j = 0; j < 30; j++) {
        if ((i >> j) & 1) {
          if (cur != 0) {
            res.push_back(
                make_pair('+', make_pair(cur, (long long)x * (1ll << j))));
          }
          cur += (long long)x * (1ll << j);
        }
      }
      for (pair<long long, long long> f : ans) {
        res.push_back(make_pair('^', f));
      }
      if (!add(1, false)) {
        break;
      }
    }
  }
  cout << res.size() << '\n';
  for (auto g : res) {
    printf("%lld %c %lld\n", g.second.first, g.first, g.second.second);
  }
}