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 md = 1e9 + 7;
const int N = 200005;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, l, r;
  long long a[3][200005], p[3];
  cin >> n >> l >> r;
  for (int i = 0; i < 3; i++) p[i] = 0;
  int l1 = l, r1 = r;
  if (r - l < 105) {
    while (l1 <= r1) {
      p[l1 % 3]++;
      l1++;
    }
  } else {
    while (!(l1 % 3 == 1)) {
      p[l1 % 3]++;
      l1++;
    }
    while (!(r1 % 3 == 0)) {
      p[r1 % 3]++;
      r1--;
    }
    int k = r1 - l1 + 1;
    if (k > 0) {
      for (int i = 0; i < 3; i++) p[i] += (k / 3);
    }
  }
  for (int i = 0; i < 3; i++) a[i][0] = p[i];
  for (int i = 1; i < n; i++) {
    a[0][i] = (a[0][i - 1] * p[0]) % md + (a[1][i - 1] * p[2]) % md +
              (a[2][i - 1] * p[1]) % md;
    a[1][i] = (a[0][i - 1] * p[1]) % md + (a[1][i - 1] * p[0]) % md +
              (a[2][i - 1] * p[2]) % md;
    a[2][i] = (a[0][i - 1] * p[2]) % md + (a[1][i - 1] * p[1]) % md +
              (a[2][i - 1] * p[0]) % md;
    for (int j = 0; j < 3; j++) a[j][i] %= md;
  }
  cout << a[0][n - 1];
  return 0;
}
