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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CF1427E extends PrintWriter {
	CF1427E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1427E o = new CF1427E(); o.main(); o.flush();
	}

	static final int M = 100000;
	long[] aa = new long[M];
	long[] bb = new long[M];
	char[] op = new char[M];
	int cnt;
	void add(long a, long b) {
		aa[cnt] = a; bb[cnt] = b; op[cnt] = '+'; cnt++;
		//println(a + " + " + b + " = " + (a + b));
	}
	void xor(long a, long b) {
		aa[cnt] = a; bb[cnt] = b; op[cnt] = '^'; cnt++;
		//println(a + " ^ " + b + " = " + (a ^ b));
	}
	void main() {
		long x = sc.nextInt(), a, b;
		if ((x & 2) == 2) {
			add(x, x); a = x + x;
			xor(x, a); x = x ^ a;
		}
		int k = 0;
		while (1 << k + 1 <= x)
			k++;
		a = x;
		for (int h = 0; h < k; h++) {
			add(a, a); a = a + a;
		}
		add(a, x); b = a + x;
		xor(b, a); b = b ^ a;
		xor(b, x); b = b ^ x;
		while (k > 0) {
			a = x;
			for (int h = 0; h < k; h++) {
				add(a, a); a = a + a;
				if ((a & b) == b) {
					xor(a, b); a = a ^ b;
				}
			}
			if ((x & a) == a) {
				xor(x, a); x = x ^ a;
				if (x == 1)
					break;
			}
			b = a;
			k--;
		}
		println(cnt);
		for (int h = 0; h < cnt; h++)
			println(aa[h] + " " + op[h] + " " + bb[h]);
	}
}