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 long long MOD = 1e9 + 7;
long long n, l, r, o, z, t, dp[200090][5];
long long Dp(int i, int sum) {
  if (i == n) {
    return (sum == 0);
  }
  if (dp[i][sum] != -1) return dp[i][sum];
  long long ret1 = 0;
  ret1 += o * Dp(i + 1, (sum + 1) % 3);
  ret1 %= MOD;
  ret1 += t * Dp(i + 1, (sum + 2) % 3);
  ret1 %= MOD;
  ret1 += z * Dp(i + 1, (sum + 3) % 3);
  ret1 %= MOD;
  return dp[i][sum] = ret1;
}
int main() {
  cin >> n >> l >> r;
  long long len = r - l + 1;
  if (len % 3 == 1) {
    if (l % 3 == 0) z++;
    if (l % 3 == 1) o++;
    if (l % 3 == 2) t++;
  }
  if (len % 3 == 2) {
    if (l % 3 == 0) {
      z++;
      o++;
    }
    if (l % 3 == 1) {
      t++;
      o++;
    }
    if (l % 3 == 2) {
      t++;
      z++;
    }
  }
  t += len / 3;
  o += len / 3;
  z += len / 3;
  memset(dp, -1, sizeof dp);
  cout << Dp(0, 0);
  return 0;
}
