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;
using ll = long long;
using ii = pair<int, int>;
const int N = 200005;
const int mod = 1e9 + 7;
int n, l, r;
int m[4];
ll dp[N][4];
ll f(int id, int sum) {
  if (id == n) return sum == 0;
  ll &ans = dp[id][sum];
  if (ans != -1) return ans;
  ans = (f(id + 1, sum) * m[0]) % mod;
  ans += (f(id + 1, (sum + 1) % 3) * m[1]) % mod;
  ans += (f(id + 1, (sum + 2) % 3) * m[2]) % mod;
  ans %= mod;
  return ans;
}
int main() {
  scanf("%d %d %d", &n, &l, &r);
  m[0] = r / 3 - (l - 1) / 3;
  m[1] = (r + 2) / 3 - (l + 1) / 3;
  m[2] = (r + 1) / 3 - l / 3;
  memset(dp, -1, sizeof dp);
  printf("%lld\n", f(0, 0));
  return 0;
}
