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;
int main() {
  long long n, m, q, t, k, i, j, l, r, minn, z0, z1, z2, x1, x2;
  int a[3];
  int aa[3];
  long long z[3];
  long long pp;
  string s;
  cin >> n >> l >> r;
  k = r - l + 1;
  z[0] = k / 3;
  z[1] = k / 3;
  z[2] = k / 3;
  x1 = l % 3;
  x2 = r % 3;
  if (x1 == x2)
    z[x1]++;
  else {
    if ((x1 == 0) and (x2 == 1)) {
      z[0]++;
      z[1]++;
    }
    if ((x1 == 1) and (x2 == 2)) {
      z[1]++;
      z[2]++;
    }
    if ((x1 == 2) and (x2 == 0)) {
      z[2]++;
      z[0]++;
    }
  }
  aa[0] = z[0];
  aa[1] = z[1];
  aa[2] = z[2];
  for (i = 2; i <= n; i++) {
    a[0] = (aa[0] * z[0] + aa[1] * z[2] + aa[2] * z[1]) % 1000000007;
    a[1] = (aa[0] * z[1] + aa[1] * z[0] + aa[2] * z[2]) % 1000000007;
    a[2] = (aa[0] * z[2] + aa[1] * z[1] + aa[2] * z[0]) % 1000000007;
    aa[0] = a[0];
    aa[1] = a[1];
    aa[2] = a[2];
  }
  cout << aa[0];
}
