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].
from math import *
from collections import *
def mul(a,b):
    return ((a%mod)*(b%mod))%mod

mod = int(1e9 + 7)
n,l,r = map(int,input().split())
l -= 1
s = [0,0,0]
s[0] = r//3-l//3;
s[1] = (r+2)//3-(l+2)//3;
s[2] = (r+1)//3-(l+1)//3;

dp = [[0,0,0] for i in range(n+1)]
dp[1][0] = s[0]
dp[1][1] = s[1]
dp[1][2] = s[2]
for i in range(1,n+1):
    for j in range(3):
        for k in range(3):
            if (k+j)%3 == 1:
                dp[i][1] = (dp[i][1]%mod + mul(dp[i-1][j],s[k])%mod)%mod
            elif (k+j)%3 == 2:
                dp[i][2] = (dp[i][2]%mod + mul(dp[i-1][j],s[k])%mod)%mod
            elif (k+j)%3 == 0:
                dp[i][0] = (dp[i][0]%mod + mul(dp[i-1][j],s[k])%mod)%mod
    #print(dp[i])

print(dp[n][0]%mod)