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 PYTHON3):

Solution:
def pr(ans):
    print(len(ans))
    for i in ans:
        if i[0] == 1:
            print(i[1], "+", i[2])
        else:
            print(i[1], "^", i[2])
def fi2(st):
    return pow(2,st - 1)
def colv(num, fi, mod):
    return pow(num, fi - 1, mod)
step2 = set()
for i in range(64):
    step2.add(pow(2,i))
a = int(input())
ans = []
x = a
k = 0
for i in range(100):
    ans.append([1, x, a])
    y = x + a
    if y in step2:
        k = y
        break
    ans.append([2, x, y])
    x = y ^ x
    if x in step2:
        k = x
        break
if k == 1:
    pr(ans)
    exit()
stk = 0
kcopy = k
while kcopy % 2 == 0:
    stk += 1
    kcopy //= 2
step = fi2(stk)
num = colv(a, step, k) - 1
tosteps = [a]
for i in range(1, 64):
    ans.append([1, tosteps[-1], tosteps[-1]])
    t = tosteps[-1] * 2
    if t >= k:
        ans.append([2, t, k])
        t -= k
    tosteps.append(t)
an = a
for i in range(64):
    if num % 2 == 1:
        ans.append([1, an, tosteps[i]])
        an += tosteps[i]
        if an > k:
            ans.append([2, an, k])
            an -= k
    num //= 2
pr(ans)