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 int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = res * x;
    x = x * x;
    n = n / 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    long long int n, l, r, i, x = 0, y = 0, z = 0, j;
    cin >> n >> l >> r;
    for (i = l; i <= r; i++) {
      if (i % 3 == 1) {
        x++;
      }
      if (i % 3 == 2) {
        y++;
      }
      if (i % 3 == 0) {
        break;
      }
    }
    for (j = r; j > i; j--) {
      if (j % 3 == 1) x++;
      if (j % 3 == 2) y++;
      if (j % 3 == 0) break;
    }
    if (j >= i) {
      x = x + (j - i) / 3;
      y = y + (j - i) / 3;
      z = z + (j - i) / 3 + 1;
    }
    x = x % 1000000007;
    y = y % 1000000007;
    z = z % 1000000007;
    long long int dp[3][n + 1];
    memset(dp, 0, sizeof(dp));
    dp[0][1] = z;
    dp[1][1] = x;
    dp[2][1] = y;
    for (i = 2; i <= n; i++) {
      dp[0][i] =
          ((z * dp[0][i - 1]) % 1000000007 + (dp[1][i - 1] * y) % 1000000007 +
           (dp[2][i - 1] * x) % 1000000007) %
          1000000007;
      dp[1][i] =
          ((dp[0][i - 1] * x) % 1000000007 + (dp[2][i - 1] * y) % 1000000007 +
           (dp[1][i - 1] * z) % 1000000007) %
          1000000007;
      dp[2][i] =
          ((dp[0][i - 1] * y) % 1000000007 + (dp[1][i - 1] * x) % 1000000007 +
           (dp[2][i - 1] * z) % 1000000007) %
          1000000007;
    }
    cout << dp[0][n] % 1000000007;
  }
}
