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;
int const INF = (int)1e9 + 1e3;
long long const INFL = (long long)1e18 + 1e6;
mt19937 tw(chrono::high_resolution_clock::now().time_since_epoch().count());
uniform_int_distribution<long long> ll_distr;
long long rnd(long long a, long long b) {
  return ll_distr(tw) % (b - a + 1) + a;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
pair<long long, long long> gcd2(long long a, long long b) {
  if (b == 0) {
    return {1, 0};
  }
  long long c = a / b;
  auto p = gcd2(b, a % b);
  return {p.second, p.first - c * p.second};
}
void solve() {
  set<long long> se;
  vector<pair<pair<long long, long long>, long long>> ans;
  long long x;
  cin >> x;
  se.insert(x);
  auto add = [&](long long a, long long b) {
    if (se.count(a + b) == 0) {
      ans.push_back(make_pair(make_pair(a, b), 0));
      se.insert(a + b);
    }
    return;
  };
  auto rox = [&](long long a, long long b) {
    if (se.count(a ^ b) == 0) {
      ans.push_back(make_pair(make_pair(a, b), 1));
      se.insert(a ^ b);
    }
    return;
  };
  auto mul = [&](long long a, long long b) {
    long long ans = 0;
    while (b) {
      if (b & 1) {
        if (ans != 0) add(ans, a);
        ans += a;
      }
      b /= 2;
      if (b) add(a, a);
      a += a;
    }
    return;
  };
  for (long long i = 0;; i++) {
    long long a = (i * x) ^ ((i + 1) * x);
    if (gcd(a, x) == 1) {
      mul(x, i);
      mul(x, i + 1);
      rox(x * i, x * i + x);
      auto p = gcd2(x, a);
      p.first = abs(p.first);
      p.second = abs(p.second);
      mul(x, p.first);
      mul(a, p.second);
      long long q = x * p.first;
      long long w = a * p.second;
      if (q > w) swap(q, w);
      if (q & 1) {
        add(q, x);
        add(w, x);
        q += x;
        w += x;
      }
      rox(q, w);
      break;
    }
  }
  cout << ans.size() << "\n";
  for (auto p : ans) {
    cout << p.first.first << " ";
    if (p.second)
      cout << "^ ";
    else
      cout << "+ ";
    cout << p.first.second << "\n";
  }
}
int main() {
  cout << setprecision(15) << fixed;
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int test_count = 1;
  for (int test = 1; test <= test_count; ++test) {
    solve();
  }
}