Alice and Bob are playing a game on a line with n cells. There are n cells labeled from 1 through n. For each i from 1 to n-1, cells i and i+1 are adjacent.

Alice initially has a token on some cell on the line, and Bob tries to guess where it is. 

Bob guesses a sequence of line cell numbers x_1, x_2, …, x_k in order. In the i-th question, Bob asks Alice if her token is currently on cell x_i. That is, Alice can answer either "YES" or "NO" to each Bob's question.

At most one time in this process, before or after answering a question, Alice is allowed to move her token from her current cell to some adjacent cell. Alice acted in such a way that she was able to answer "NO" to all of Bob's questions.

Note that Alice can even move her token before answering the first question or after answering the last question. Alice can also choose to not move at all.

You are given n and Bob's questions x_1, …, x_k. You would like to count the number of scenarios that let Alice answer "NO" to all of Bob's questions. 

Let (a,b) denote a scenario where Alice starts at cell a and ends at cell b. Two scenarios (a_i, b_i) and (a_j, b_j) are different if a_i ≠ a_j or b_i ≠ b_j.

Input

The first line contains two integers n and k (1 ≤ n,k ≤ 10^5) — the number of cells and the number of questions Bob asked.

The second line contains k integers x_1, x_2, …, x_k (1 ≤ x_i ≤ n) — Bob's questions.

Output

Print a single integer, the number of scenarios that let Alice answer "NO" to all of Bob's questions.

Examples

Input


5 3
5 1 4


Output


9


Input


4 8
1 2 3 4 4 3 2 1


Output


0


Input


100000 1
42


Output


299997

Note

The notation (i,j) denotes a scenario where Alice starts at cell i and ends at cell j.

In the first example, the valid scenarios are (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 5). For example, (3,4) is valid since Alice can start at cell 3, stay there for the first three questions, then move to cell 4 after the last question. 

(4,5) is valid since Alice can start at cell 4, stay there for the first question, the move to cell 5 for the next two questions. Note that (4,5) is only counted once, even though there are different questions that Alice can choose to do the move, but remember, we only count each pair of starting and ending positions once.

In the second example, Alice has no valid scenarios.

In the last example, all (i,j) where |i-j| ≤ 1 except for (42, 42) are valid scenarios.
"""
    Satwik_Tiwari ;) .
    30th july , 2020  - Thursday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase

from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================
#fast I/O region
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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))

def solve():
    n,k = sep()
    x = lis()
    pos = [[] for i in range (n+1)]
    # print(pos)
    for i in range(k):
        # print(x[i])
        pos[x[i]].append(i+1)
    for i in range(n+1):
        pos[i] = sorted(pos[i])

    ans =0
    for i in range(1,n+1):
        if(len(pos[i]) == 0):
            if(i==1 or i==n):
                ans+=2
            else:
                ans+=3
        else:
            if(i<n):
                if(len(pos[i+1])==0 or pos[i][0]>pos[i+1][len(pos[i+1])-1]):
                    ans+=1
            if(i>1):
                if(len(pos[i-1])==0 or pos[i][0]>pos[i-1][len(pos[i-1])-1]):
                    ans+=1
    print(ans)




testcase(1)
# testcase(int(inp()))












