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;
const long long maxn = 1e6 + 10, lg = 62;
long long n, k, A[maxn];
vector<pair<pair<long long, long long>, char> > v;
void mul(long long a, long long c) {
  long long ans = 0;
  while (c > 0) {
    if (c & 1ll) {
      if (ans > 0ll) {
        v.push_back(make_pair(make_pair(a, ans), '+'));
      }
      ans += a;
    }
    if (c > 1) {
      v.push_back(make_pair(make_pair(a, a), '+'));
    }
    a *= 2ll;
    c /= 2ll;
  }
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long findB(long long x, long long y, long long tar) {
  for (long long b = 0; b < x; b++) {
    long long r = (y * b) % x;
    if (r == tar % x) {
      return b;
    }
  }
}
long long getY(long long x) {
  for (long long i = 2; i < 2 * maxn; i++) {
    if (gcd((x * i) ^ x, x) == 1) {
      return i;
    }
  }
}
void build(long long x, long long y, long long c) {
  long long b1 = findB(x, y, c);
  long long a1 = (c - y * b1) / x;
  mul(y, b1);
  mul(x, a1);
  if (a1 > 0 && b1 > 0) {
    v.push_back(make_pair(make_pair(x * a1, y * b1), '+'));
  }
}
void solve(long long x) {
  v.clear();
  long long yp = getY(x);
  mul(x, yp);
  v.push_back(make_pair(make_pair(x, x * yp), '^'));
  long long y = ((x * yp) ^ x);
  build(x, y, (1ll << lg));
  build(x, y, (1ll << lg) + 1);
  v.push_back(make_pair(make_pair((1ll << lg), (1ll << lg) + 1), '^'));
}
int32_t main() {
  long long x;
  cin >> x;
  solve(x);
  cout << v.size() << endl;
  for (pair<pair<long long, long long>, char> p : v) {
    cout << p.first.first << ' ' << p.second << ' ' << p.first.second << endl;
  }
}