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 long long M = 1000000007;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long x;
  cin >> x;
  vector<tuple<long long, char, long long>> A;
  while (x != 1) {
    long long bit = (long long)ceil(log2(x + 1));
    long long now = x;
    for (long long _ = 0; _ < bit - 1; _++) {
      A.push_back({now, '+', now});
      now += now;
    }
    long long chel = now;
    A.push_back({now, '+', x});
    A.push_back({now, '^', x});
    long long l1 = now + x, l2 = now ^ x;
    A.push_back({l1, '+', l2});
    long long h = l1 + l2;
    A.push_back({now, '+', now});
    now += now;
    A.push_back({now, '^', h});
    now ^= h;
    A.push_back({now, '^', x + x});
    now ^= (x + x);
    while (now + now <= 1e12) {
      A.push_back({now, '+', now});
      now += now;
    }
    for (long long i = 50; i >= bit; i--) {
      if ((chel >> i) & 1) {
        A.push_back({chel, '^', (1LL << i)});
        chel ^= (1LL << i);
      }
    }
    A.push_back({x, '^', chel});
    x ^= chel;
  }
  cout << A.size() << '\n';
  for (auto c : A) {
    cout << get<0>(c) << ' ' << get<1>(c) << ' ' << get<2>(c) << '\n';
  }
}