Ayoub had an array a of integers of size n and this array had two interesting properties: 

  * All the integers in the array were between l and r (inclusive). 
  * The sum of all the elements was divisible by 3. 



Unfortunately, Ayoub has lost his array, but he remembers the size of the array n and the numbers l and r, so he asked you to find the number of ways to restore the array. 

Since the answer could be very large, print it modulo 10^9 + 7 (i.e. the remainder when dividing by 10^9 + 7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.

Input

The first and only line contains three integers n, l and r (1 ≤ n ≤ 2 ⋅ 10^5 , 1 ≤ l ≤ r ≤ 10^9) — the size of the lost array and the range of numbers in the array.

Output

Print the remainder when dividing by 10^9 + 7 the number of ways to restore the array.

Examples

Input


2 1 3


Output


3


Input


3 2 2


Output


1


Input


9 9 99


Output


711426616

Note

In the first example, the possible arrays are : [1,2], [2,1], [3, 3].

In the second example, the only possible array is [2, 2, 2].
import math
def func():
    global x1,x2,x3
    # if(n==0):
    #     if(x==0):
    #         return x3 
    #     elif(x==1):
    #         return x1 
    #     else:
    #         return x2
    # if(x==0):
    #     return x3*func(n-1,0) + x1*func(n-1,2) + x2*func(n-1,1)
    # elif(x==1):
    #     return x3*func(n-1,1) + x1*func(n-1,0) + x2*func(n-1,2)
    # elif(x==2):
    #     return x3*func(n-1,2) + x1*func(n-1,1) + x2*func(n-1,0)
    x1=x1% p 
    x2=x2 % p 
    x3=x3% p
    for i in range(n):
        for j in range(3):
            if(i==0):
                if(j==0):
                    dp[i][j]=x3 
                elif(j==1):
                    dp[i][j]=x1 
                else:
                    dp[i][j]=x2 
            else:
                if(j==0):
                    dp[i][j]=x3*dp[i-1][0] + x1*dp[i-1][2] + x2*dp[i-1][1]
                elif(j==1):
                    dp[i][j]=x3*dp[i-1][1] + x1*dp[i-1][0] + x2*dp[i-1][2]
                else:
                    dp[i][j]=x3*dp[i-1][2] + x1*dp[i-1][1] + x2*dp[i-1][0]
                dp[i][j]=dp[i][j]%p
    return dp[n-1][0]
n,l,r=tuple(map(int,input().split()))
p = 10**9 + 7
if(l==r):
    if(n*l%3==0):
        print(1)
    else:
        print(0)
else:
    # if(r==l+1):
    #     if(r%3==0):
    #         x3=1
    #         x2=1
    #         x1=0
    #     elif(l%3==0):
    #         x3=1
    #         x1=1
    #         x2=0
    # elif(r==l+2):
    #     x1=1
    #     x2=1
    #     x3=1
    # else:
        x3 = math.floor(r/3) - math.ceil(l/3) + 1
        x1=x3-1
        x2=x3-1
        if(l%3==1):
            x1+=1
            x2+=1
        elif(l%3==2):
            x2+=1
        if(r%3==1):
            x1+=1
        elif(r%3==2):
            x1+=1
            x2+=1
        dp=[[-1,-1,-1] for i in range(n)]
        print(func()%p)