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;
template <typename T1, typename T2>
inline void chkmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chkmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
vector<pair<char, pair<ll, ll> > > res;
set<ll> have;
vector<ll> haves;
vector<ll> fd;
ll add(ll a, ll b) {
  if (have.find(a + b) != have.end()) {
    return a + b;
  }
  if (a + b >= 1000000000000000000LL) {
    return a + b;
  }
  res.push_back(make_pair('+', make_pair(a, b)));
  fd.push_back(a + b);
  have.insert(a + b);
  haves.push_back(a + b);
  return a + b;
}
ll xorM(ll a, ll b) {
  if (have.find((a ^ b)) != have.end()) {
    return (a ^ b);
  }
  if ((a ^ b) >= 1000000000000000000LL) {
    return (a ^ b);
  }
  have.insert((a ^ b));
  res.push_back(make_pair('^', make_pair(a, b)));
  fd.push_back(a ^ b);
  haves.push_back(a ^ b);
  return a ^ b;
}
struct BitwiseGauss {
  int B;
  vector<ll> bb;
  BitwiseGauss(int _B) {
    B = _B;
    bb.resize(B);
  }
  bool insert1(ll x) {
    vector<pair<ll, ll> > kek;
    for (int i = B - 1; i >= 0; --i) {
      if (((x >> i) & 1)) {
        if (bb[i]) {
          x ^= bb[i];
        } else {
          return true;
        }
      }
    }
    return false;
  }
  void insert(ll x) {
    vector<pair<ll, ll> > kek;
    for (int i = B - 1; i >= 0; --i) {
      if (((x >> i) & 1)) {
        if (bb[i]) {
          kek.push_back(make_pair(x, bb[i]));
          x ^= bb[i];
        } else {
          for (auto z : kek) {
            xorM(z.first, z.second);
          }
          bb[i] = x;
          break;
        }
      }
    }
  }
  bool can1(ll x) {
    for (int i = B - 1; i >= 1; --i) {
      if (((x >> i) & 1)) {
        if (bb[i]) {
          x ^= bb[i];
        }
      }
    }
    if (x == 0) {
      if (bb[0]) {
        x ^= bb[0];
      }
    }
    return x == 1;
  }
  bool can(ll x) {
    for (int i = B - 1; i >= 1; --i) {
      if (((x >> i) & 1)) {
        if (bb[i]) {
          xorM(x, bb[i]);
          x ^= bb[i];
        }
      }
    }
    if (x == 0) {
      if (bb[0]) {
        xorM(x, bb[0]);
        x ^= bb[0];
      }
    }
    return x == 1;
  }
  void clear() { fill(bb.begin(), bb.end(), 0); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  BitwiseGauss kek(61);
  ll x;
  cin >> x;
  have.clear();
  haves.clear();
  have.insert(x);
  haves.push_back(x);
  kek.insert(x);
  res.clear();
  mt19937 as;
  while ((!kek.can1(x) && !kek.bb[0])) {
    fd.clear();
    while (fd.empty()) {
      ll f = haves[as() % (int)(haves).size()];
      ll f1 = haves[as() % (int)(haves).size()];
      if (as() & 1) {
        ll vl = f + f1;
        if (vl <= 1000000000000000000LL) {
          if (have.find(vl) == have.end()) {
            if (kek.insert1(vl)) {
              ll op = add(f, f1);
            }
          }
        }
      } else {
        ll vl = f ^ f1;
        if (vl <= 1000000000000000000LL) {
          if (have.find(vl) == have.end()) {
            if (kek.insert1(vl)) {
              ll op = xorM(f, f1);
            }
          }
        }
      }
    }
    while (!fd.empty() && (!kek.can1(x) && !kek.bb[0])) {
      auto x = fd.back();
      fd.pop_back();
      if (kek.insert1(x)) {
        kek.insert(x);
      }
    }
  }
  assert(kek.can(x));
  assert((int)(res).size() <= 100000);
  cout << (int)(res).size() << '\n';
  for (auto x : res) {
    cout << x.second.first << ' ' << x.first << ' ' << x.second.second << '\n';
  }
}