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].
import sys
stdin = sys.stdin
 
sys.setrecursionlimit(10**5) 
 
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

n,l,r = li()
MOD = 10**9+7

st = l%3
lg = (r-l+1)//3
mods = [lg]*3
for i in range((r-l+1)-3*lg):
    mods[st] += 1
    st += 1
    st %= 3
    
dp = [[0]*3 for _ in range(n+1)]
dp[0][0] = 1

for i in range(n):
    dp[i+1][0] = dp[i][0]*mods[0]%MOD + dp[i][1]*mods[2]%MOD + dp[i][2]*mods[1]%MOD
    dp[i+1][1] = dp[i][0]*mods[1]%MOD + dp[i][1]*mods[0]%MOD + dp[i][2]*mods[2]%MOD
    dp[i+1][2] = dp[i][0]*mods[2]%MOD + dp[i][1]*mods[1]%MOD + dp[i][2]*mods[0]%MOD

    dp[i+1][0] %= MOD
    dp[i+1][1] %= MOD
    dp[i+1][2] %= MOD
    
print(dp[-1][0])