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].
m = 10**(9)+7
def f(l,r):
    if l%3 == 0 and r%3 == 0:
        z = (r-l)//3+1
        o = (r-l)//3
        t = (r-l)//3
    elif l%3 == 0 and r%3 == 1:
        z = o = (r-l)//3+1
        t = z-1
    elif l%3 == 0 and r%3 == 2:
        z = t = o = (r-l)//3+1
    elif l%3 == 1 and r%3 == 0:
        z = t = o = (r-l)//3+1
    elif l%3 == 1 and r%3 == 1:
        z =  t = (r-l)//3
        o = (r-l)//3+1
    elif l%3 == 1 and r%3 == 2:
        z = (r-l)//3
        t = o = (r-l)//3+1
    elif l%3 == 2 and r%3 == 0:
        z = t = (r-l)//3+1
        o = (r-l)//3
    elif l%3 == 2 and r%3 == 1:
        z = t = o = (r-l)//3+1
    elif l%3 == 2 and r%3 == 2:
        o = z = (r-l)//3
        t = (r-l)//3+1
    return([z,o,t])
n,l,r = map(int,input().split())
l = f(l,r)
dp = [[0 for i in range(n+1)] for j in range(3)]
for j in range(3):
    dp[0][1] = l[0]
    dp[1][1] = l[1]
    dp[2][1] = l[2]
for i in range(2,n+1):
    dp[0][i] = ((dp[0][i-1]%m)*(l[0]%m)+(dp[1][i-1]%m)*(l[2]%m)+(dp[2][i-1]%m)*(l[1]%m))%m
    dp[1][i] = ((dp[0][i-1]%m)*(l[1]%m)+(dp[1][i-1]%m)*(l[0]%m)+(dp[2][i-1]%m)*(l[2]%m))%m
    dp[2][i] = ((dp[0][i-1]%m)*(l[2]%m)+(dp[1][i-1]%m)*(l[1]%m) +(dp[2][i-1]%m)*(l[0]%m))%m
print(dp[0][n]%m)
    
    