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

dp = [[0]*3 for i in range(200005)]

n, l, r = map(int, input().split())
l3, r3 = l%3, r%3
k = r-l+1

if l3 == r3:	
    zero = k  // 3
    one = k  // 3
    two = k  // 3;
    if ( l3 == 0 ):
        zero+=1
    elif ( l3 == 1 ):
        one+=1
    else:
        two+=1

elif ( l3 == 0 and r3 == 1 ):
    zero, one, two = k  // 3 + 1,k  // 3 + 1,k  // 3
elif ( l3 == 1 and r3 == 2 ):
    zero, one, two = k  // 3, k  // 3 + 1, k  // 3 + 1
elif ( l3 == 2 and r3 == 0 ):
    zero,one,two = k  // 3 + 1, k  // 3, k  // 3 + 1
else:
    zero,one,two = k  // 3, k  // 3, k  // 3;


dp[1][0], dp[1][1], dp[1][2] = zero, one, two
#print(zero, one, two)

for i in range(2, n+1):
    dp[i][0], dp[i][1], dp[i][2] =(
        (dp[i-1][0]*zero + dp[i-1][1]*two + dp[i-1][2]*one)%mod,
        (dp[i-1][0]*one + dp[i-1][1]*zero + dp[i-1][2]*two)%mod,
        (dp[i-1][0]*two + dp[i-1][1]*one + dp[i-1][2]*zero)%mod
    )

print(dp[n][0])