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 lint = long long;
class Gaussian {
 public:
  mt19937 rnd;
  vector<lint> basis;
  Gaussian()
      : rnd(chrono::high_resolution_clock::now().time_since_epoch().count()) {}
  bool Insert(lint x) {
    for (auto i : basis) {
      if ((x ^ i) < x) {
        x ^= i;
      }
    }
    if (x != 0) {
      basis.emplace_back(x);
      for (int i = (int)basis.size() - 1; i > 0; i--) {
        if (basis[i - 1] < basis[i]) {
          swap(basis[i - 1], basis[i]);
        }
      }
    }
    return x != 0;
  }
  lint GetRandom() {
    lint res = 0;
    for (auto i : basis) {
      if (rnd() & 1) {
        res ^= i;
      }
    }
    return res;
  }
  bool IsIndependent(lint x) {
    for (auto i : basis) {
      if ((i ^ x) < x) {
        x ^= i;
      }
    }
    return x != 0;
  }
  void Backtrack(lint x, vector<array<lint, 3>> &op, bool getrand) {
    lint res = 0;
    for (auto i : basis) {
      if ((x ^ i) < x) {
        if (getrand) {
          op.push_back({1, res, i});
          res ^= i;
        } else {
          op.push_back({1, i, x});
        }
        x ^= i;
      }
    }
  }
};
void Main() {
  lint X;
  cin >> X;
  vector<array<lint, 3>> ans;
  ans.push_back({1, X, X});
  Gaussian G;
  G.Insert(X);
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  vector<lint> A = {X};
  while (G.IsIndependent(1)) {
    int u = rnd() % A.size();
    int v = rnd() % A.size();
    if (A[u] + A[v] < (lint)4e18 && G.IsIndependent(A[u] + A[v])) {
      ans.push_back({0, A[u], A[v]});
      A.emplace_back(A[u] + A[v]);
      if (G.IsIndependent(A[u] + A[v])) {
        G.Backtrack(A[u] + A[v], ans, false);
        G.Insert(A[u] + A[v]);
      }
    }
  }
  assert(count(begin(G.basis), end(G.basis), 1) > 0);
  assert(ans.size() <= 100000);
  cout << ans.size() << "\n";
  for (auto i : ans) {
    if (i[0] == 0) {
      cout << i[1] << " + " << i[2] << "\n";
    } else {
      cout << i[1] << " ^ " << i[2] << "\n";
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  while (T--) {
    Main();
  }
  return 0;
}