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 MAX = (int)1e5 + 55;
const int INF = (int)1e9 + 77;
const int MOD = (int)1e9 + 7;
const double PI = 2 * acos(0.0);
const double EPS = 1e-7;
const int N = (int)2e5 + 5;
int dp[N][3];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, l, r;
  cin >> n >> l >> r;
  long long mod0 = ((r / 3) - ((l - 1) / 3));
  long long mod1 = ((r + 2) / 3) - ((l + 1) / 3);
  long long mod2 = ((r + 1) / 3) - ((l) / 3);
  dp[0][0] = mod0;
  dp[0][1] = mod1;
  dp[0][2] = mod2;
  for (int i = 1; i < n; ++i) {
    long long val0 = mod0 * dp[i - 1][0];
    val0 %= MOD;
    val0 += mod1 * dp[i - 1][2];
    val0 %= MOD;
    val0 += mod2 * dp[i - 1][1];
    val0 %= MOD;
    dp[i][0] = val0;
    long long val1 = mod0 * dp[i - 1][1];
    val1 %= MOD;
    val1 += mod1 * dp[i - 1][0];
    val1 %= MOD;
    val1 += mod2 * dp[i - 1][2];
    val1 %= MOD;
    dp[i][1] = val1;
    long long val2 = mod0 * dp[i - 1][2];
    val2 %= MOD;
    val2 += mod2 * dp[i - 1][0];
    val2 %= MOD;
    val2 += mod1 * dp[i - 1][1];
    val2 %= MOD;
    dp[i][2] = val2;
  }
  cout << dp[n - 1][0];
  return 0;
}
