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>
long long dp[3][200005];
int main() {
  int n;
  long long l, r;
  scanf("%d %lld %lld", &n, &l, &r);
  int ml, mr;
  ml = l % 3;
  mr = r % 3;
  long long n0, n1, n2;
  long long mn = r - l + 1 - (3 - ml) - (mr + 1);
  n0 = mn / 3;
  n1 = n0;
  n2 = n0;
  if (ml == 0) {
    n0++;
    n1++;
    n2++;
  } else if (ml == 1) {
    n1++;
    n2++;
  } else {
    n2++;
  }
  if (mr == 0) {
    n0++;
  } else if (mr == 1) {
    n0++;
    n1++;
  } else {
    n0++;
    n1++;
    n2++;
  }
  dp[0][1] = n0;
  dp[1][1] = n1;
  dp[2][1] = n2;
  int i;
  for (i = 2; i <= n; i++) {
    dp[0][i] = (dp[0][i - 1] * n0 + dp[1][i - 1] * n2 + dp[2][i - 1] * n1) %
               1000000007;
    dp[1][i] = (dp[0][i - 1] * n1 + dp[1][i - 1] * n0 + dp[2][i - 1] * n2) %
               1000000007;
    dp[2][i] = (dp[0][i - 1] * n2 + dp[1][i - 1] * n1 + dp[2][i - 1] * n0) %
               1000000007;
  }
  printf("%lld\n", dp[0][n]);
}
