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, l, r, nol, odin, dva, nol1, odin1, dva1;
  cin >> n >> l >> r;
  vector<long long> a(3);
  if (l % 3 == 0) {
    if (r % 3 == 0) {
      nol = (r - l) / 3 + 1;
      dva = odin = nol - 1;
    }
    if (r % 3 == 1) {
      nol = odin = (r - l) / 3 + 1;
      dva = nol - 1;
    }
    if (r % 3 == 2) {
      nol = odin = dva = (r - l) / 3 + 1;
    }
  }
  if (l % 3 == 1) {
    if (r % 3 == 2) {
      nol = (r - l) / 3;
      dva = odin = nol + 1;
    }
    if (r % 3 == 1) {
      nol = dva = (r - l) / 3;
      odin = nol + 1;
    }
    if (r % 3 == 0) {
      nol = odin = dva = (r - l) / 3 + 1;
    }
  }
  if (l % 3 == 2) {
    if (r % 3 == 0) {
      nol = dva = (r - l) / 3 + 1;
      odin = nol - 1;
    }
    if (r % 3 == 2) {
      nol = odin = (r - l) / 3;
      dva = nol + 1;
    }
    if (r % 3 == 1) {
      nol = odin = dva = (r - l) / 3 + 1;
    }
  }
  a[0] = nol;
  a[1] = odin;
  a[2] = dva;
  for (int i = 1; i < n; i++) {
    nol1 = nol;
    dva1 = dva;
    odin1 = odin;
    nol = (nol1 * a[0] + odin1 * a[2] + dva1 * a[1]) % (long long)(1e9 + 7);
    odin = (odin1 * a[0] + dva1 * a[2] + nol1 * a[1]) % (long long)(1e9 + 7);
    dva = (dva1 * a[0] + odin1 * a[1] + nol1 * a[2]) % (long long)(1e9 + 7);
  }
  cout << nol;
  return 0;
}
