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, mn, mx = map(int, input("").split(" "))
MOD = 1000000007

def cntMod(l, r, m):
  lb = 0
  ub = 0
  if (l % 3 == m):
    lb = l
  else:
    if m > l % 3:
      lb = l + (m - l % 3)
    else:
      lb = l + (m + 3 - l % 3)
  if r % 3 == m:
    ub = r
  else:
    if m < r % 3:
      ub = r - (r % 3 - m)
    else:
      ub = r - (r % 3 - m + 3)
  if lb > ub:
    return 0
  else:
    return (ub - lb) / 3 + 1

mod0 = int(cntMod(mn, mx, 0)) % MOD
mod1 = int(cntMod(mn, mx, 1)) % MOD
mod2 = int(cntMod(mn, mx, 2)) % MOD
dp = []
for i in range(n + 1):
  dp.append([None, None, None])

for i in range(n + 1):
  for j in range(3):
    if i == 0:
      if j == 0:
        dp[i][j] = 1
      else:
        dp[i][j] = 0
    else:
      a = (mod0*dp[i - 1][j]) % MOD
      b = (mod1*dp[i - 1][(j + 1) % 3]) % MOD
      c = (mod2*dp[i - 1][(j + 2) % 3]) % MOD
      d = (a + b) % MOD
      dp[i][j] = (d + c) % MOD

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