Rebel spy Heidi has just obtained the plans for the Death Star from the Empire and, now on her way to safety, she is trying to break the encryption of the plans (of course they are encrypted – the Empire may be evil, but it is not stupid!). The encryption has several levels of security, and here is how the first one looks.

Heidi is presented with a screen that shows her a sequence of integers A and a positive integer p. She knows that the encryption code is a single number S, which is defined as follows:

Define the score of X to be the sum of the elements of X modulo p.

Heidi is given a sequence A that consists of N integers, and also given an integer p. She needs to split A into 2 parts such that: 

  * Each part contains at least 1 element of A, and each part consists of contiguous elements of A. 
  * The two parts do not overlap. 
  * The total sum S of the scores of those two parts is maximized. This is the encryption code. 



Output the sum S, which is the encryption code.

Input

The first line of the input contains two space-separated integer N and p (2 ≤ N ≤ 100 000, 2 ≤ p ≤ 10 000) – the number of elements in A, and the modulo for computing scores, respectively.

The second line contains N space-separated integers which are the elements of A. Each integer is from the interval [1, 1 000 000].

Output

Output the number S as described in the problem statement.

Examples

Input

4 10
3 4 7 2


Output

16


Input

10 12
16 3 24 13 9 8 7 5 12 12


Output

13

Note

In the first example, the score is maximized if the input sequence is split into two parts as (3, 4), (7, 2). It gives the total score of <image>.

In the second example, the score is maximized if the first part consists of the first three elements, and the second part consists of the rest. Then, the score is <image>.
#Mamma don't raises quitter.................................................
from collections import deque as de
import math
import re
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb
from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack() 

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

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

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))
    return prime_factors

def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

 
# Function to get product of digits
def getProduct(n):
 
    product = 1
 
    while (n != 0):
        product = product * (n % 10)
        n = n // 10
 
    return product

# function to count consecutive duplicate element in an array
def dupconscount(nums):
    element = []
    freque = []
    if not nums:
        return element
    running_count = 1
    for i in range(len(nums)-1):
        if nums[i] == nums[i+1]:
            running_count += 1
        else:
            freque.append(running_count)
            element.append(nums[i])
            running_count = 1
    freque.append(running_count)
    element.append(nums[i+1])
    return element,freque
        

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 

#ceil  function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def ceildiv(x,y): 
    return (x+y-1)//y 
  
def di():return map(int, input().split())
def ii():return int(input())
def li():return list(map(int, input().split()))
def si():return list(map(str, input()))
def indic():
    dic = {}
    for index, value in enumerate(input().split()):
        dic[int(index)+1] = int(value)
    return dic

#Here we go......................
#concentration and mental toughness are margins of victory
n,p=di()
a=li()
fir=a[0]
sec=sum(a)-fir
ans=(fir%p)+(sec%p)
for i in range(1,n):
    fir+=a[i]
    sec-=a[i]
    temp=(fir%p)+(sec%p)
    if temp >ans:
        ans=temp
print(ans)

                    
                    
            
            
    
    

   
        





    
            
        
        
    
    

    
    
    
        


        
    




    





    





    
    
    
    
    

          
            
    

    
    
    
    
    
    
        
        
            
            
            
    
        
        
        

                

        
    



        
            

        

        
    
        
         





                
        
            
        
            

        
        

    

    
    
            
        
    

             
        


    
        

            

        
    
    

    
        
    
    

            
            
            

        
    
        
    
    
    
    

    
            
        
        
    
        
        
        


                
                

    


        
        
    
    
    
        
    
        
    




    

                
    
        


        
    

            
        
    

                
                
                
        
        
        

            

        





                    
                
            
        

            
    
        
    
    

    
        
    
 

    
        


    
        

