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>
#pragma GCC optimize("O2")
using namespace std;
const double PI = atan(1.0) * 4;
const int INF = 1e9;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, -1, 0, 1};
const int maximum = numeric_limits<int>::max();
const int minimum = numeric_limits<int>::min();
const int mod = 1e9 + 7;
int binpow(int a, int b, int m) {
  a %= m;
  int res = 1;
  while (b > 0) {
    if (b & 1) res = 1ll * res * a % m;
    a = 1ll * a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  T = 1;
  while (T--) {
    long long n, l, r;
    cin >> n >> l >> r;
    vector<long long> num(3);
    num[0] = r / 3 - (l - 1) / 3;
    num[1] = (r + 2) / 3 - (l + 1) / 3;
    num[2] = (r + 1) / 3 - l / 3;
    vector<vector<long long>> dp(n + 1, vector<long long>(3));
    dp[0][0] = 1;
    dp[0][1] = 1;
    dp[0][2] = 1;
    dp[1][0] = num[0];
    dp[1][1] = num[1];
    dp[1][2] = num[2];
    for (int i = 2; i <= n; i++) {
      for (int j = 0; j < 3; j++) {
        if (j == 0) {
          dp[i][j] = dp[i - 1][0] * num[0] % mod + dp[i - 1][1] * num[2] % mod +
                     dp[i - 1][2] * num[1] % mod;
          dp[i][j] %= mod;
        } else if (j == 1) {
          dp[i][j] = dp[i - 1][0] * num[1] % mod + dp[i - 1][1] * num[0] % mod +
                     dp[i - 1][2] * num[2] % mod;
          dp[i][j] %= mod;
        } else if (j == 2) {
          dp[i][j] = dp[i - 1][0] * num[2] % mod + dp[i - 1][1] * num[1] % mod +
                     dp[i - 1][2] * num[0] % mod;
          dp[i][j] %= mod;
        }
      }
    }
    cout << dp[n][0] << "\n";
  }
  return 0;
}
