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;
long long dp[200000][3];
long long n, l, r;
long long r0, r1, r2;
long long mod = 1000000007;
int main() {
  std::cin >> n >> l >> r;
  r0 = (r / 3) - (l - 1) / 3;
  r1 = (r + 1) / 3 - l / 3;
  r2 = (r + 2) / 3 - (l + 1) / 3;
  dp[0][0] = r0;
  dp[0][1] = r1;
  dp[0][2] = r2;
  for (int i = 1; i < n; i++) {
    dp[i][0] = dp[i - 1][0] * r0 + dp[i - 1][1] * r2 + dp[i - 1][2] * r1;
    dp[i][0] %= mod;
    dp[i][1] = dp[i - 1][0] * r1 + dp[i - 1][1] * r0 + dp[i - 1][2] * r2;
    dp[i][1] %= mod;
    dp[i][2] = dp[i - 1][0] * r2 + dp[i - 1][1] * r1 + dp[i - 1][2] * r0;
    dp[i][2] %= mod;
  }
  std::cout << dp[n - 1][0];
}
