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;
int main() {
  long long n, i, j, l, r, cnt0, cnt1, cnt2, tot;
  cin >> n >> l >> r;
  tot = r - l + 1;
  long long dp[n][3];
  cnt0 = cnt1 = cnt2 = tot / 3;
  if (tot % 3 == 1) {
    if (l % 3 == 1)
      cnt1++;
    else if (l % 3 == 2)
      cnt2++;
    else
      cnt0++;
  } else if (tot % 3 == 2) {
    if (l % 3 == 1) {
      cnt1++;
      cnt2++;
    } else if (l % 3 == 2) {
      cnt2++;
      cnt0++;
    } else {
      cnt0++;
      cnt1++;
    }
  }
  dp[0][0] = cnt0;
  dp[0][1] = cnt1;
  dp[0][2] = cnt2;
  for (i = 1; i < n; i++) {
    dp[i][0] = ((dp[i - 1][0]) % 1000000007 * cnt0) % 1000000007 +
               ((dp[i - 1][1]) % 1000000007 * cnt2) % 1000000007 +
               ((dp[i - 1][2]) % 1000000007 * cnt1) % 1000000007;
    dp[i][0] %= 1000000007;
    dp[i][1] = (dp[i - 1][0] * cnt1) % 1000000007 +
               (dp[i - 1][1] * cnt0) % 1000000007 +
               (dp[i - 1][2] * cnt2) % 1000000007;
    dp[i][1] %= 1000000007;
    dp[i][2] = (dp[i - 1][0] * cnt2) % 1000000007 +
               (dp[i - 1][1] * cnt1) % 1000000007 +
               (dp[i - 1][2] * cnt0) % 1000000007;
    dp[i][2] %= 1000000007;
  }
  cout << dp[n - 1][0];
}
