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].
n,l,r=map(int,input().split())
t=(r+1-l)//3
c0,c1,c2=t,t,t
if t!=(r+1-l)/3:
    if (r+1-l)%3==1:
        if l%3==1:
            c1+=1
        elif l%3==0:
            c0+=1
        else:
            c2+=1
    if (r+1-l)%3==2:
        if l%3==1:
            c1+=1
            c2+=1
        elif l%3==0:
            c0+=1
            c1+=1
        else:
            c2+=1
            c0+=1
dp=[[0]*3 for i in range(n)]
dp[0][0]=c0
dp[0][1]=c1
dp[0][2]=c2
for i in range(1,n):
    dp[i][0]=((dp[i-1][0]*c0)+(dp[i-1][1]*c2)+(dp[i-1][2]*c1))%(10**9+7)
    dp[i][1]=((dp[i-1][0]*c1)+(dp[i-1][1]*c0)+(dp[i-1][2]*c2))%(10**9+7)
    dp[i][2]=((dp[i-1][0]*c2)+(dp[i-1][2]*c0)+(dp[i-1][1]*c1))%(10**9+7)
print(dp[-1][0]%(10**9+7))