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;
const int MAX = 3e5 + 11;
const int MAXN = 5e7 + 1;
const int BITMAX = 18;
const int INF = 1e9 + 2;
const long long LLINF = 1e17 + 3;
const int MOD = 1e9 + 7;
const long double Pi = acos(-1);
int n, l, r;
int dp[MAX][4];
int solve(int id, int md) {
  if (id == n) return md == 0;
  int &ret = dp[id][md];
  if (ret != -1) return ret;
  ret = 0;
  int tmp1 = md + r;
  int r1 = (r - l + 3) / 3;
  int tmp2 = md + r - 1;
  int r2 = (r - l + 2) / 3;
  int tmp3 = md + r - 2;
  int r3 = (r - l + 1) / 3;
  ret = (1ll * (r1)*solve(id + 1, tmp1 % 3) + ret) % MOD;
  ret = (1ll * (r2)*solve(id + 1, tmp2 % 3) + ret) % MOD;
  ret = (1ll * (r3)*solve(id + 1, tmp3 % 3) + ret) % MOD;
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &l);
  scanf("%d", &r);
  memset(dp, -1, sizeof dp);
  printf("%d\n", solve(0, 0));
  return 0;
}
