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 = list(map(int,input().split(' ')))
dp1 = []
dp2 = []
dp0 = []
a = r//3 - l//3

dp0.append(a)
dp1.append(a)
dp2.append(a)
N = 10**9+7
if(l==r):
    if (l*n)%3==0:
        print('1')
    else:
        print('0')
else:
    if l%3==0:
        if r%3==0:
            dp0[0]+=1
        elif r%3==1:
            dp1[0]+=1
            dp0[0]+=1
        elif r%3==2:
            dp2[0]+=1
            dp1[0]+=1
            dp0[0]+=1
    elif l%3==1:
        if r%3==1:
            dp1[0]+=1
        elif r%3==2:
            dp2[0]+=1
            dp1[0]+=1
    
    elif l%3==2:
        if r%3==0:
            dp1[0]-=1
        elif r%3==2:
            dp2[0]+=1
    
    for i in range(1,n):
        dp0.append((dp0[i-1]*dp0[0]+dp1[i-1]*dp2[0]+dp2[i-1]*dp1[0])%N)
        dp1.append((dp0[i-1]*dp1[0]+dp1[i-1]*dp0[0]+dp2[i-1]*dp2[0])%N)
        dp2.append((dp0[i-1]*dp2[0]+dp1[i-1]*dp1[0]+dp2[i-1]*dp0[0])%N)
    
    print(dp0[n-1])
        
        


    
    