Problem: Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.

Initially you have a sequence of n integers: 1, 2, ..., n. In a single step, you can pick two of them, let's denote them a and b, erase them from the sequence, and append to the sequence either a + b, or a - b, or a × b.

After n - 1 steps there is only one number left. Can you make this number equal to 24?

Input

The first line contains a single integer n (1 ≤ n ≤ 105).

Output

If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).

If there is a way to obtain 24 as the result number, in the following n - 1 lines print the required operations an operation per line. Each operation should be in form: "a op b = c". Where a and b are the numbers you've picked at this operation; op is either "+", or "-", or "*"; c is the result of corresponding operation. Note, that the absolute value of c mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.

If there are multiple valid answers, you may print any of them.

Examples

Input

1


Output

NO


Input

8


Output

YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
enum OP { ADD, MINUS1, MINUS2, MULT, NOP };
struct Arith {
  long long res;
  OP op;
  Arith *lval, *rval;
  Arith(const long long &rs, const OP &o, Arith *const lv, Arith *const rv)
      : res(rs), op(o), lval(lv), rval(rv) {}
  Arith(const long long &rs) : res(rs), op(NOP), lval(NULL), rval(NULL) {}
};
vector<Arith *> num;
bool found = false;
Arith *sol = NULL;
void solve() {
  if (num.size() == 1) {
    if (num[0]->res == 24L) {
      found = true;
      sol = num[0];
    }
    return;
  }
  for (int i = 0; i < num.size(); ++i) {
    for (int j = i + 1; j < num.size(); ++j) {
      Arith *ll = num[i], *rr = num[j];
      num.erase(num.begin() + i);
      num.erase(num.begin() + (j - 1));
      for (int k = 0; k < NOP; ++k) {
        OP op = static_cast<OP>(k);
        Arith *c = NULL;
        switch (op) {
          case ADD:
            c = new Arith(ll->res + rr->res, ADD, ll, rr);
            break;
          case MINUS1:
            c = new Arith(ll->res - rr->res, MINUS1, ll, rr);
            break;
          case MINUS2:
            c = new Arith(rr->res - ll->res, MINUS2, rr, ll);
            break;
          case MULT:
            c = new Arith(ll->res * rr->res, MULT, ll, rr);
            break;
          default:
            break;
        }
        num.push_back(c);
        solve();
        if (found) return;
        num.pop_back();
      }
      num.insert(num.begin() + i, ll);
      num.insert(num.begin() + j, rr);
    }
  }
}
void printFormula(const Arith *const node) {
  if (node->lval->op != NOP) {
    printFormula(node->lval);
  }
  if (node->rval->op != NOP) {
    printFormula(node->rval);
  }
  cout << node->lval->res;
  switch (node->op) {
    case ADD:
      cout << " + ";
      break;
    case MINUS1:
      cout << " - ";
      break;
    case MINUS2:
      cout << " - ";
      break;
    case MULT:
      cout << " * ";
      break;
    default:
      break;
  }
  cout << node->rval->res << " = " << node->res << endl;
}
void padBefore(const long long &s, const long long &n) {
  for (long long i = s; i <= n - 1; i += 2) {
    cout << (i + 1) << " - " << i << " = 1" << endl;
  }
}
void padAfter(const long long &n) {
  long long t = n / 2 - 2;
  while (t--) {
    cout << "24 * 1 = 24" << endl;
  }
}
int main() {
  long long n;
  cin >> n;
  if (n <= 3) {
    cout << "NO" << endl;
    return 0;
  }
  long long cnt = (n % 2 == 0 ? 4 : 5);
  for (long long i = 1; i <= cnt; ++i) {
    Arith *node = new Arith(i);
    num.push_back(node);
  }
  solve();
  if (found) {
    cout << "YES" << endl;
    padBefore(cnt + 1, n);
    printFormula(sol);
    padAfter(n);
  } else {
    cout << "NO" << endl;
  }
  return 0;
}