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;
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
const long long maxn = 1000010;
const long long mod = 1000000007;
const long double eps = 1e-9;
const long long inf = ((1ll << 31ll) - 1ll);
const long long INF = 1ll * mod * mod;
const long double pi = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long qpow(long long b, long long e) {
  if (!e) return 1;
  if (e & 1) return qpow(b, e - 1) * b % mod;
  long long pwur = qpow(b, e >> 1);
  return pwur * pwur % mod;
}
long long modinv(long long x) { return qpow(x, mod - 2); }
vector<long long> vx;
long long cnt;
struct pr {
  long long a, b, o;
};
vector<pr> prnt;
long long g(long long val) {
  long long res = 0;
  while (val % 2 == 0) {
    val >>= 1;
    res++;
  }
  return res;
}
void solve(long long x) {
  if (x == 1) return;
  set<long long> second, vx;
  second.insert(x);
  second.insert(mod);
  for (long long i = 1; i <= 5; i++) {
    set<long long> w;
    w.insert(mod);
    for (auto i : second)
      for (auto j : second) {
        if (i == mod || j == mod) continue;
        if (*second.lower_bound(i + j) != i + j &&
            *w.lower_bound(i + j) != i + j && i + j <= 5 * x) {
          w.insert(i + j);
          prnt.push_back({i, j, 0});
          if ((i + j) % 2) vx.insert(i + j);
        }
        if (*second.lower_bound((i ^ j)) != (i ^ j) &&
            *w.lower_bound((i ^ j)) != (i ^ j) && (i ^ j) <= 5 * x) {
          w.insert(i ^ j);
          prnt.push_back({i, j, 1});
          if ((i ^ j) % 2) vx.insert((i ^ j));
        }
      }
    for (auto i : w) {
      second.insert(i);
      if (i == 1) return;
    }
  }
  vector<pair<long long, long long> > vy;
  for (auto i : vx)
    if (i < 2 * x)
      vy.push_back({__builtin_popcountll(i) - 63 - __builtin_clzll(i), i});
  if (vy.empty()) {
    for (auto i : vx)
      if (i < 3 * x)
        vy.push_back({__builtin_popcountll(i) - 63 - __builtin_clzll(i), i});
  }
  if (vy.empty()) {
    for (auto i : vx)
      if (i < 6 * x)
        vy.push_back({__builtin_popcountll(i) - 63 - __builtin_clzll(i), i});
  }
  sort(vy.begin(), vy.end());
  return solve(vy[rand() % min((long long)vy.size(), 10ll)].second);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(0);
  srand(time(NULL));
  long long n;
  cin >> n;
  solve(n);
  while (prnt.size() > 100000) solve(n);
  cout << prnt.size() << '\n';
  for (auto i : prnt) {
    if (i.o == 0) cout << i.a << " + " << i.b << '\n';
    if (i.o == 1) cout << i.a << " ^ " << i.b << '\n';
  }
  return 0;
}