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())
mod = 1000000007

nl = l+(3-l%3)%3
nr = r-r%3
q0 = q1 = q2 =0
if nr>=nl:
    q0 = q1 = q2=(nr-nl)//3
    for i in range(r, nr, -1):
        if i % 3 == 0:
            q0 += 1
        elif i % 3 == 1:
            q1 += 1
        else:
            q2 += 1

for i in range(l,min(nl,r)+1):
    if i%3==0:
        q0+=1
    elif i%3==1:
        q1+=1
    else:
        q2+=1

q = [q0,q1,q2]

dp = [q]+[[0,0,0]for i in range(n-1)]
for l in range(1,n):
    for m in range(3):
        ans = 0
        for i in range(3):
            ans+=((dp[l-1][i]%mod)*(q[(m+3-i)%3])%mod)%mod
        dp[l][m] = ans%mod

print(dp[-1][0])