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;
const int N = 100010;
int p[N];
int pos[N];
int q[N];
void solve() {
  long long x;
  cin >> x;
  vector<pair<long long, vector<long long> > > basis;
  for (long long i = 1; i <= 10000000; i++) {
    long long y = (x * i);
    for (auto &cur : basis) {
      long long z = cur.first;
      long long nxt = (y ^ z);
      if (nxt < y) {
        y = nxt;
      }
    }
    if (y) {
      y = x * i;
      set<long long> st = {i};
      for (auto cur : basis) {
        long long z = cur.first;
        long long nxt = (y ^ z);
        if (nxt < y) {
          y = nxt;
          for (long long j : cur.second) {
            if (st.find(j) == st.end())
              st.insert(j);
            else
              st.erase(j);
          }
        }
      }
      vector<long long> v;
      for (long long j : st) v.push_back(j);
      basis.push_back({y, v});
    }
    if (y == 1) {
      break;
    }
  }
  vector<long long> v = basis.back().second;
  vector<pair<int, pair<long long, long long> > > ans;
  for (long long i = 0; i < 30; i++) {
    ans.push_back({1, {x * (1ll << i), x * (1ll << i)}});
  }
  for (long long y : v) {
    long long cur = 0ll;
    for (long long i = 0; i < 30; i++) {
      if (y & (1ll << i)) {
        if (cur) ans.push_back({1, {cur, x * (1ll << i)}});
        cur += x * (1ll << i);
      }
    }
  }
  long long cur = v[0] * x;
  for (int i = 1; i < v.size(); i++) {
    ans.push_back({2, {cur, v[i] * x}});
    cur ^= v[i] * x;
  }
  cout << ans.size() << '\n';
  for (auto cur : ans) {
    if (cur.first == 1)
      cout << cur.second.first << " + " << cur.second.second << '\n';
    else
      cout << cur.second.first << " ^ " << cur.second.second << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}