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:
N=1000
q=[0,0,1,1]+[0,1]*(N//2)
for i in range(3,N,2):
    if q[i]:
        for j in range(3*i,N,2*i):q[j]=0
p=[]
for i in range(2,N):
    if q[i]:p.append(i)
def phi(n):
    r=1
    for i in p:
        if n%i<1:
            r*=i-1
            n//=i
            while n%i<1:
                n//=i
                r*=i
    r*=max(n-1,1)
    return r
n=int(input())
o=[]
vn=bin(n)[3:]
kn=n
for i in range(len(vn)):o.append(f'{kn} + {kn}');kn*=2
o.append(f'{kn} ^ {n}')
a=kn^n
b=pow(a,phi(n)-1,n)
if b%2<1:b+=n
v=bin(b)[3:]
k=a
for i in range(len(v)):o.append(f'{k} + {k}');k*=2
j=k
k//=2
for i in range(len(v)):
    if v[i]=='1':
        o.append(f'{j} + {k}')
        j+=k
    k//=2
c=j//n
v=bin(c)[3:]
k=kn
for i in range(len(v)-len(vn)):o.append(f'{k} + {k}');k*=2
l=k
k//=2
for i in range(len(v)):
    if v[i]=='1':
        o.append(f'{l} + {k}')
        l+=k
    k//=2
o.append(f'{j} ^ {l}')
print(len(o))
print('\n'.join(o))