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 os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

n,r,l = map(int,input().split())

mod = 1000000007
l0 = l%3
q_l = l//3

if l0 == 0 :
	l1 = q_l +1
	l2 = q_l
	l3 = q_l
elif l0 == 1 :
	l1 = q_l + 1
	l2 = q_l + 1
	l3 = q_l
else :
	l1 = q_l + 1
	l2 = q_l + 1
	l3 = q_l +1

r0 = (r-1)%3
r_l = (r-1)//3

if r0 == 0 :
	r1 = r_l +1
	r2 = r_l
	r3 = r_l
elif r0 == 1 :
	r1 = r_l + 1
	r2 = r_l + 1
	r3 = r_l
else :
	r1 = r_l + 1
	r2 = r_l + 1
	r3 = r_l +1


k1 = l1 - r1
k2 = l2 - r2
k3 = l3 - r3



l1 = [0 for i in range(n)]
l2 = [0 for i in range(n)]
l3 = [0 for i in range(n)]

l1[0] = k1
l2[0] = k2
l3[0] = k3

for i in range(1,n) :
	l1[i] = (k1*l1[i-1] + k3*l2[i-1] + k2*l3[i-1] )%mod
	l2[i] = (k2*l1[i-1] + k1*l2[i-1] + k3*l3[i-1] )%mod
	l3[i] = (k3*l1[i-1] + k2*l2[i-1] + k1*l3[i-1] )%mod


print(l1[n-1]%mod)

