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 int64_t mod = 1e9 + 7;
const int64_t M = 2e5 + 5;
int64_t n, l, r, dp[M][3], L[3], R[3];
int64_t rem0, rem1, rem2;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> r;
  L[l % 3] = l, R[r % 3] = r;
  int64_t tL = l, tR = r;
  l++, r--;
  if (l <= tR && r >= tL) L[l % 3] = l, R[r % 3] = r;
  l++, r--;
  if (l <= tR && r >= tL) L[l % 3] = l, R[r % 3] = r;
  if (R[0] && L[0]) rem0 = (R[0] - L[0]) / 3 + 1;
  if (R[1] && L[1]) rem1 = (R[1] - L[1]) / 3 + 1;
  if (R[2] && L[2]) rem2 = (R[2] - L[2]) / 3 + 1;
  dp[1][0] = rem0, dp[1][1] = rem1, dp[1][2] = rem2;
  for (int64_t i = 2; i <= n; i++) {
    dp[i][0] =
        (rem0 * dp[i - 1][0] + rem1 * dp[i - 1][2] + rem2 * dp[i - 1][1]) % mod;
    dp[i][1] =
        (rem0 * dp[i - 1][1] + rem1 * dp[i - 1][0] + rem2 * dp[i - 1][2]) % mod;
    dp[i][2] =
        (rem0 * dp[i - 1][2] + rem1 * dp[i - 1][1] + rem2 * dp[i - 1][0]) % mod;
  }
  cout << dp[n][0];
  return 0;
}
