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("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using datas = pair<ll, ll>;
using ddatas = pair<long double, long double>;
using tdata = pair<ll, datas>;
using vec = vector<ll>;
using mat = vector<vec>;
using pvec = vector<datas>;
using pmat = vector<pvec>;
const ll mod = 1000000007;
const ll inf = 1LL << 60;
const long double PI = acos(-1);
const long double eps = 1e-9;
template <class T, class E>
ostream& operator<<(ostream& os, pair<T, E>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class T>
inline bool chmax(T& a, T b) {
  bool x = a < b;
  if (x) a = b;
  return x;
}
template <class T>
inline bool chmin(T& a, T b) {
  bool x = a > b;
  if (x) a = b;
  return x;
}
void startupcpp() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
long double distance(ddatas x, ddatas y) {
  long double a = x.first - y.first, b = x.second - y.second;
  return sqrt(a * a + b * b);
}
ll modinv(ll a, ll m = mod) {
  ll b = m, u = 1, v = 0, t;
  while (b) {
    t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return (u + m) % m;
}
ll moddevide(ll a, ll b) { return (a * modinv(b)) % mod; }
vec modncrlistp, modncrlistm;
ll modncr(ll n, ll r) {
  if (n < r) return 0;
  ll i, size = modncrlistp.size();
  if (size <= n) {
    modncrlistp.resize(n + 1);
    modncrlistm.resize(n + 1);
    if (!size) {
      modncrlistp[0] = modncrlistm[0] = 1;
      size++;
    }
    for (i = size; i < (ll)n + 1; ++i) {
      modncrlistp[i] = modncrlistp[i - 1] * i % mod;
      modncrlistm[i] = modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n] * modncrlistm[r] % mod * modncrlistm[n - r] % mod;
}
ll modpow(ll a, ll n, ll m = mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % m;
    a = a * a % m;
    n >>= 1;
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (!b) return abs(a);
  return (a % b == 0) ? abs(b) : gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll countdigits(ll n) {
  ll ans = 0;
  while (n) {
    n /= 2;
    ans++;
  }
  return ans;
}
ll sumdigits(ll n) {
  ll ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
struct edge {
  ll a, b;
  string op;
};
vector<edge> v;
ll pr1(ll a, ll b) {
  v.push_back({a, b, " + "});
  return a + b;
}
ll pr2(ll a, ll b) {
  v.push_back({a, b, " ^ "});
  return a ^ b;
}
void solve(ll x) {
  if (x == 1) return;
  if (x & 2) {
    pr1(x, x);
    solve(pr2(x << 1, x));
    return;
  }
  ll y = x, K = countdigits(x) - 1;
  while (K--) y = pr1(y, y);
  ll t = pr2(x, y);
  ll z = pr2(pr1(x, y), t);
  while (t > x) {
    z = pr1(z, z);
    if ((t ^ z) < t) t = pr2(t, z);
  }
  solve(t);
}
int main() {
  ll N;
  cin >> N;
  solve(N);
  cout << v.size() << "\n";
  for (auto x : v) {
    cout << x.a << x.op << x.b << "\n";
  }
}