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 b, a, mins, l, r, c, tc, ar[200005][4], mods = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> l >> r;
  l--;
  long long ls = l % 3, l1 = l / 3, l2 = l / 3, l3 = l / 3;
  if (ls > 0) {
    l1++;
  }
  if (ls > 1) {
    l2++;
  }
  long long rs = r % 3, r1 = r / 3, r2 = r / 3, r3 = r / 3;
  if (rs > 0) {
    r1++;
  }
  if (rs > 1) {
    r2++;
  }
  long long j1 = r1 - l1, j2 = r2 - l2, j3 = r3 - l3;
  ar[0][0] = 1;
  for (int k = 1; k <= a; k++) {
    ar[k][1] += (ar[k - 1][0] * j1) % mods;
    ar[k][1] %= mods;
    ar[k][1] += (ar[k - 1][1] * j3) % mods;
    ar[k][1] %= mods;
    ar[k][1] += (ar[k - 1][2] * j2) % mods;
    ar[k][1] %= mods;
    ar[k][2] += (ar[k - 1][0] * j2) % mods;
    ar[k][2] %= mods;
    ar[k][2] += (ar[k - 1][1] * j1) % mods;
    ar[k][2] %= mods;
    ar[k][2] += (ar[k - 1][2] * j3) % mods;
    ar[k][2] %= mods;
    ar[k][0] += (ar[k - 1][0] * j3) % mods;
    ar[k][0] %= mods;
    ar[k][0] += (ar[k - 1][1] * j2) % mods;
    ar[k][0] %= mods;
    ar[k][0] += (ar[k - 1][2] * j1) % mods;
    ar[k][0] %= mods;
  }
  cout << ar[a][0] << '\n';
}
