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].
MOD = 1000000007
n,l,r = map(int,input().split())
dp = []
for i in range(n):
    dp.append([0,0,0])
Bw = r - l + 1
X = int(Bw//3)
Zero = X
One = X
Two = X
Rem = Bw % 3
if(Rem == 1):
    if(l%3 == 0):
        Zero+=1
    elif(l%3 == 1):
        One += 1
    else:
        Two += 1
elif(Rem == 2):
    if(l%3 == 0):
        Zero+=1
        One += 1
    elif(l%3 == 1):
        One += 1
        Two += 1
    else:
        Two += 1
        Zero += 1
dp[0][0] = Zero % MOD
dp[0][1] = One % MOD
dp[0][2] = Two % MOD
for i in range(1,n):
    dp[i][0] = (dp[i-1][0] * Zero + dp[i-1][1] * Two +  dp[i-1][2] * One) % MOD
    dp[i][1] = (dp[i-1][0] * One + dp[i-1][1] * Zero + dp[i-1][2] * Two) % MOD
    dp[i][2] = (dp[i-1][0] * Two + dp[i-1][1] * One +  dp[i-1][2] * Zero) % MOD
print(dp[n-1][0])