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())
dp=[[0]*3 for _ in range(n+5)]

mod = 10**9+7
rcnt=[0]*3

ll=1+l//3
rr=r//3
for i in range(3):
    rcnt[i]+=rr-ll

if l%3<=0:
    rcnt[0]+=1
if l%3<=1:
    rcnt[1]+=1
rcnt[2]+=1

if r%3>=0:
    rcnt[0]+=1
if r%3>=1:
    rcnt[1]+=1
if r%3==2:
    rcnt[2]+=1

for i in range(3):
    dp[1][i]=rcnt[i]%mod

for i in range(2,n+1):
    x = [dp[i-1][0] * rcnt[0] % mod, dp[i-1][2] * rcnt[1] % mod, dp[i-1][1] * rcnt[2] % mod]
    y = [dp[i-1][0] * rcnt[1] % mod, dp[i-1][1] * rcnt[0] % mod, dp[i-1][2] * rcnt[2] % mod]
    z = [dp[i-1][0] * rcnt[2] % mod, dp[i-1][1] * rcnt[1] % mod, dp[i-1][2] * rcnt[0] % mod]
    dp[i][0] = sum(x) % mod
    dp[i][1] = sum(y) % mod
    dp[i][2] = sum(z) % mod

print(dp[n][0])