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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 998244353;
const ll root = 62;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
ll modInv(ll a) { return binpow(a, MOD - 2); }
const double PI = atan(1);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
ll x;
vector<array<ll, 3>> ans;
ll add(ll a, ll b) {
  ans.push_back({a, 1, b});
  return a + b;
}
ll xorr(ll a, ll b) {
  ans.push_back({a, 0, b});
  return a ^ b;
}
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin.tie(0)->sync_with_stdio(0);
  cin >> x;
  if (x % 4 == 3) {
    add(x, add(x, x));
    x *= 3;
  }
  int highbit = 0;
  ll sv = x;
  while (sv) {
    sv /= 2;
    highbit++;
  }
  ll curr = x;
  for (auto i = (0); i < (highbit - 1); ++i) {
    curr = add(curr, curr);
  }
  ll temp = curr;
  temp = add(curr, x);
  temp = xorr(temp, xorr(curr, x));
  while (temp > 1) {
    if (x > temp) {
      x = xorr(x, temp);
    }
    sv = x;
    for (auto i = (0); i < (highbit - 1); ++i) {
      sv = add(sv, sv);
      if (sv > temp) {
        sv = xorr(sv, temp);
      }
    }
    highbit--;
    temp = sv;
  }
  cout << (ans.size()) << "\n";
  for (auto p : ans) {
    for (auto i = (0); i < (3); ++i) {
      if (i == 1) {
        cout << (p[i] == 1 ? " + " : " ^ ");
      } else {
        cout << p[i];
      }
    }
    cout << "\n";
  }
  return 0;
}