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,raw_input().split())
dp=[]
for i in range(0,n+4):
    t=[]
    for j in range(0,3):
        if i==0:
            t.append(1)
        else:
            t.append(0)
    dp.append(t)
mod=10**9+7
def solve(r):
    if r%3==0:
        v1=r/3
        v2=r/3
        v3=r/3
    elif r%3==1:
        v1=r/3
        v2=r/3+1
        v3=r/3
    else:
        v1=r/3
        v2=r/3+1
        v3=r/3+1
    return v1,v2,v3
tu1=solve(r)
tu2=solve(l-1)
#print tu1,tu2
val0=tu1[0]-tu2[0]
val1=tu1[1]-tu2[1]
val2=tu1[2]-tu2[2]
dp[1][1]=val1
dp[1][2]=val2
dp[1][0]=val0
for i in range(1,n+1):
    dp[i+1][0]=dp[i][0]*(val0)+dp[i][1]*val2+dp[i][2]*val1
    dp[i+1][0]%=mod
    dp[i+1][1]=dp[i][1]*val0+dp[i][0]*val1+dp[i][2]*val2
    dp[i+1][1]%=mod
    dp[i+1][2]=dp[i][1]*val1+dp[i][0]*val2+dp[i][2]*val0
    dp[i+1][2]%=mod
#print dp
print dp[n][0]%mod


