human
Give me a Python program to find any such graph! given the following:
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!

When building the graph, he needs four conditions to be satisfied: 

  * It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops. 
  * The number of vertices must be exactly n — a number he selected. This number is not necessarily prime. 
  * The total number of edges must be prime. 
  * The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime. 


Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4. 


Note that the graph can be disconnected.

Please help Bob to find any such graph!

Input

The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.

Output

If there is no graph satisfying the conditions, print a single line containing the integer -1.

Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.

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

Note that the graph can be disconnected.

Examples

Input

4

Output

5
1 2
1 3
2 3
2 4
3 4

Input

8

Output

13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8

Note

The first example was described in the statement.

In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is...
#----
assistant
#Here is Python program to find any such graph!
.
--------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
#-------------------game starts now-----------------------------------------------------
def prime(n) : 
    # Corner cases 
    if (n <= 1) : 
        return False
    if (n <= 3) : 
        return True
  
    # This is checked so that we can skip  
    # middle...