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 = list(map(int, input().split()))
dp = [[0] * 3 for i in range(n + 1)]
th = r // 3 - (l - 1) // 3
on = 0
tw = 0
iter = l
while iter % 3 != 0 and iter <= r:
    if iter % 3 == 1:
        on += 1
    elif iter % 3 == 2:
        tw += 1
    iter += 1
sector = r
while sector % 3 != 0 and sector > iter:
    if sector % 3 == 1:
        on += 1
    elif sector % 3 == 2:
        tw += 1
    sector -= 1

tw += max(th - 1, 0)
on += max(th - 1, 0)

cnt = [on, tw, th]
for i in range(1, n + 1):
    for k in range(3):
        if i == 1:
            dp[i][k] += cnt[k]
            dp[i][k] %= 1000000007
        else:
            if k == 1:
                dp[i][k] += dp[i - 1][k - 1] * on
                dp[i][k] += dp[i - 1][k] * th
                dp[i][k] += dp[i - 1][k + 1] * tw
                dp[i][k] %= 1000000007
            if k == 0:
                dp[i][k] += dp[i - 1][k + 1] * tw
                dp[i][k] += dp[i - 1][k] * th
                dp[i][k] += dp[i - 1][k + 2] * on
                dp[i][k] %= 1000000007
            if k == 2:
                dp[i][k] += dp[i - 1][k] * th
                dp[i][k] += dp[i - 1][k - 1] * on
                dp[i][k] += dp[i - 1][k - 2] * tw
                dp[i][k] %= 1000000007
print(dp[n][2])