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 mod = 1e9 + 7;
const int MAX_N = 2e5 + 5;
long long dp[MAX_N][3], cnt[3];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, l, r;
  cin >> n >> l >> r;
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < 3; i++) {
    int l2 = l, r2 = r;
    while (l2 % 3 != i && l2 <= r) l2++;
    while (r2 % 3 != i && r2 >= l) r2--;
    if (r2 >= l2) cnt[i] = (r2 - l2) / 3 + 1;
  }
  dp[1][0] = cnt[0], dp[1][1] = cnt[1], dp[1][2] = cnt[2];
  for (int i = 2; i <= n; i++) {
    dp[i][0] =
        (((dp[i - 1][0] * cnt[0]) % mod) + ((dp[i - 1][1] * cnt[2]) % mod) +
         ((dp[i - 1][2] * cnt[1]) % mod)) %
        mod;
    dp[i][1] =
        (((dp[i - 1][0] * cnt[1]) % mod) + ((dp[i - 1][1] * cnt[0]) % mod) +
         ((dp[i - 1][2] * cnt[2]) % mod)) %
        mod;
    dp[i][2] =
        (((dp[i - 1][0] * cnt[2]) % mod) + ((dp[i - 1][1] * cnt[1]) % mod) +
         ((dp[i - 1][2] * cnt[0]) % mod)) %
        mod;
  }
  cout << dp[n][0] << endl;
  return 0;
}
