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;
const int mod = 1000000007;
int n, l, r;
long long mod0, mod1, mod2, f[200100][3];
int main() {
  cin >> n >> l >> r;
  mod0 = r / 3 - l / 3;
  mod1 = mod0;
  mod2 = mod0;
  if (l % 3 == 0) mod0++;
  if (l % 3 == 2) mod1--;
  if (r % 3 == 1) mod1++;
  if (r % 3 == 2) mod1++, mod2++;
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i][0] = (f[i - 1][0] * mod0) % mod + (f[i - 1][1] * mod2) % mod +
              (f[i - 1][2] * mod1) % mod;
    f[i][0] %= mod;
    f[i][1] = (f[i - 1][1] * mod0) % mod + (f[i - 1][2] * mod2) % mod +
              (f[i - 1][0] * mod1) % mod;
    f[i][1] %= mod;
    f[i][2] = (f[i - 1][0] * mod2) % mod + (f[i - 1][1] * mod1) % mod +
              (f[i - 1][2] * mod0) % mod;
    f[i][2] %= mod;
  }
  cout << f[n][0] << endl;
  return 0;
}
