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;
using ll = long long;
using ull = unsigned long long;
using dbl = long double;
vector<tuple<long long, string, long long>> answer;
const long long maxbit = 40;
void get_degrees(long long x) {
  for (long long d = 0; d < maxbit; ++d) {
    long long t = x << d;
    answer.emplace_back(t, "+", t);
  }
}
void getmul(long long x, long long k) {
  long long sum = 0;
  for (long long bit = 0; bit < maxbit; ++bit) {
    if (k & (1LL << bit)) {
      if (sum > 0) {
        answer.emplace_back(sum, "+", x << bit);
      }
      sum += x << bit;
    }
  }
  assert(sum == k * x);
}
long long solve(long long x) {
  assert(x > 1 && (x % 2 == 1));
  long long need = 1;
  while (need <= x) need *= 2;
  need = need / 2 - 1;
  get_degrees(x);
  vector<long long> basis;
  basis.reserve(60);
  for (long long k = 1;; ++k) {
    long long num = k * x;
    for (auto vec : basis) {
      num = min(num, num ^ vec);
    }
    if (num == 0) {
      continue;
    }
    getmul(x, k);
    long long current = k * x;
    for (auto &vec : basis) {
      if ((current ^ vec) < current) {
        answer.emplace_back(vec, "^", current);
        current ^= vec;
      }
      if (vec < current) {
        swap(vec, current);
      }
    }
    assert(current > 0);
    basis.push_back(current);
    if (current > need) {
      continue;
    }
    long long t = current;
    while (t <= need) {
      answer.emplace_back(t, "+", t);
      t *= 2;
    }
    answer.emplace_back(x, "^", t);
    long long result = x ^ t;
    assert(result <= need && (result & 1));
    return result;
  }
}
void solve() {
  long long x;
  cin >> x;
  while (x > 1) {
    x = solve(x);
  }
  cout << ((ll)(answer).size()) << endl;
  for (auto [x, op, y] : answer) {
    cout << x << ' ' << op << ' ' << y << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  long long T = 1;
  for (long long i = 0; i < T; ++i) {
    solve();
  }
}