Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number <image> on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: <image>, where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109). 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
2 2


Output

8


Input

3 3
1 2 3


Output

27


Input

2 2
29 29


Output

73741817


Input

4 5
0 0 0 0


Output

1

Note

In the first sample <image>. Thus, the answer to the problem is 8.

In the second sample, <image>. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample <image>. Thus, the answer to the problem is 1.
import math
pri=pow(10,9)+7
n,x=map(int,input().split())
z=list(map(int,input().split()))
sumr=sum(z)
gcd=sumr-max(z)
t=max(z)
ans=z.copy()
for i in range(len(ans)):
    ans[i]=t-ans[i]
ans.sort()
temp=[]
count=1
from heapq import *
temp=[]
for i in range(1,len(ans)):
    if(ans[i]==ans[i-1]):
        count+=1
    else:
        temp.append([ans[i-1],count])
        count=1
temp.append([ans[-1],count])
pq=[]
from collections import *
al=defaultdict(int)

for i in range(len(temp)):
    al[temp[i][0]]=temp[i][1]
    heappush(pq,temp[i][0])

while(len(pq)):
    se=heappop(pq)
    
    r=al[se]
    c1=0
    while(r%x==0 and r>0):
        r=r//x
        c1+=1
    if(c1==0):
        ty=min(se+gcd,sumr)
        print(pow(x,ty,pri))
        exit()
    else:
        if(al[se+c1]):
            al[se+c1]+=r
        else:
            heappush(pq,se+c1)
            al[se+c1]=r
    
        
        
        
     
    
   
       
        
        
        
    
            
        

    
