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].
#include <bits/stdc++.h>
using namespace std;
int n, l, r;
long long sum[3];
long long f[200005][3];
int main() {
  scanf("%d%d%d", &n, &l, &r);
  sum[0] = sum[1] = sum[2] = r / 3;
  if (r % 3 == 1)
    sum[1]++;
  else if (r % 3 == 2)
    sum[1]++, sum[2]++;
  l--;
  for (int i = 0; i < 3; i++) sum[i] -= l / 3;
  if (l % 3 == 1)
    sum[1]--;
  else if (l % 3 == 2)
    sum[1]--, sum[2]--;
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i][1] =
        (f[i - 1][0] * sum[1] + f[i - 1][1] * sum[0] + f[i - 1][2] * sum[2]) %
        1000000007;
    f[i][0] =
        (f[i - 1][0] * sum[0] + f[i - 1][1] * sum[2] + f[i - 1][2] * sum[1]) %
        1000000007;
    f[i][2] =
        (f[i - 1][0] * sum[2] + f[i - 1][1] * sum[1] + f[i - 1][2] * sum[0]) %
        1000000007;
  }
  printf("%lld", f[n][0]);
  return 0;
}
