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 N = 2e5 + 5;
const long long inf = 1e9 + 7;
long long dp[N][3], cnt[3];
int n, l, r;
int main() {
  scanf("%d%d%d", &n, &l, &r);
  memset(dp, 0, sizeof(dp));
  memset(cnt, 0, sizeof(cnt));
  dp[0][0] = 1;
  int l_l = l / 3 * 3, r_l = r / 3 * 3;
  cnt[0] = cnt[1] = cnt[2] = (r_l - l_l) / 3;
  for (int i = l_l; i < l; i++) cnt[i - l_l]--;
  for (int i = r_l; i <= r; i++) cnt[i - r_l]++;
  for (int i = 1; i <= n; i++) {
    for (int x = 0; x < 3; x++)
      for (int y = 0; y < 3; y++) {
        dp[i][(x + y) % 3] += dp[i - 1][x] * cnt[y];
        dp[i][(x + y) % 3] %= inf;
      }
  }
  printf("%lld\n", dp[n][0]);
}
