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>
const int MOD = 1e9 + 7;
long long n, l, r;
long long ans;
long long num[10];
long long f[200010][5];
int main() {
  scanf("%lld%lld%lld", &n, &l, &r);
  long long len = r - l + 1;
  long long m = len % 3;
  long long r = len / 3;
  for (int i = 1; i <= m; ++i) ++num[(l + i - 1) % 3];
  for (int i = 0; i <= 2; ++i) num[i] += r;
  for (int i = 0; i <= 2; ++i) f[1][i] = num[i];
  for (int i = 2; i <= n; ++i) {
    f[i][0] = (f[i - 1][0] * num[0] % MOD + f[i - 1][1] * num[2] % MOD +
               f[i - 1][2] * num[1] % MOD) %
              MOD;
    f[i][1] = (f[i - 1][0] * num[1] % MOD + f[i - 1][1] * num[0] % MOD +
               f[i - 1][2] * num[2] % MOD) %
              MOD;
    f[i][2] = (f[i - 1][0] * num[2] % MOD + f[i - 1][1] * num[1] % MOD +
               f[i - 1][2] * num[0] % MOD) %
              MOD;
  }
  printf("%lld\n", f[n][0]);
  return 0;
}
