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;
long long n, l, r, m0, m1, m2, m3, m4, m5;
long long MOD = 1e9 + 7;
long long dp[200005][5];
int main() {
  cin >> n >> l >> r;
  --l;
  m0 = m1 = m2 = l / 3;
  if (l % 3 >= 1) ++m1;
  if (l % 3 == 2) ++m2;
  m3 = m4 = m5 = r / 3;
  if (r % 3 >= 1) ++m4;
  if (r % 3 == 2) ++m5;
  m0 = m3 - m0;
  m1 = m4 - m1;
  m2 = m5 - m2;
  dp[1][0] = m0;
  dp[1][1] = m1;
  dp[1][2] = m2;
  for (long long i = 2; i <= n; i++) {
    dp[i][0] =
        (dp[i - 1][0] * m0 + dp[i - 1][1] * m2 + dp[i - 1][2] * m1) % MOD;
    dp[i][1] =
        (dp[i - 1][0] * m1 + dp[i - 1][1] * m0 + dp[i - 1][2] * m2) % MOD;
    dp[i][2] =
        (dp[i - 1][0] * m2 + dp[i - 1][1] * m1 + dp[i - 1][2] * m0) % MOD;
  }
  long long ans = dp[n][0];
  cout << ans % MOD;
}
