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 po(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return result;
}
long long add(long long a, long long b) {
  long long z = ((a % 1000000007) + (b % 1000000007)) % 1000000007;
  return z % 1000000007;
}
long long mul(long long a, long long b) {
  long long z = ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  return z % 1000000007;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long n, l, r;
  cin >> n >> l >> r;
  long long n1 = 0, n2 = 0, n3 = 0;
  n1 = r / 3 - l / 3;
  if (l % 3 == 0) n1++;
  l--;
  r--;
  n2 = r / 3 - l / 3 + 1;
  if (l % 3 != 0) n2--;
  n3 = r - l + 1 - n1 - n2;
  long long dp[3][n];
  for (long long i = 0; i < 3; i++)
    for (long long j = 0; j < n; j++) dp[i][j] = 0;
  dp[0][0] = n1;
  dp[1][0] = n2;
  dp[2][0] = n3;
  for (long long i = 1; i < n; i++) {
    dp[0][i] = add(dp[0][i], mul(dp[0][i - 1], n1)) % 1000000007;
    dp[0][i] = add(dp[0][i], mul(dp[1][i - 1], n3)) % 1000000007;
    dp[0][i] = add(dp[0][i], mul(dp[2][i - 1], n2)) % 1000000007;
    dp[1][i] = add(dp[1][i], mul(dp[2][i - 1], n3)) % 1000000007;
    dp[1][i] = add(dp[1][i], mul(dp[0][i - 1], n2)) % 1000000007;
    dp[1][i] = add(dp[1][i], mul(dp[1][i - 1], n1)) % 1000000007;
    dp[2][i] = add(dp[2][i], mul(dp[1][i - 1], n2)) % 1000000007;
    dp[2][i] = add(dp[2][i], mul(dp[0][i - 1], n3)) % 1000000007;
    dp[2][i] = add(dp[2][i], mul(dp[2][i - 1], n1)) % 1000000007;
  }
  cout << dp[0][n - 1] % 1000000007;
}
