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 maxn = 200010;
const int mod = 1e9 + 7;
inline int cal(int x, int y) {
  int l = 0, r = 1e9;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (3ll * mid + y <= x)
      l = mid;
    else
      r = mid - 1;
  }
  if (y > x) return 0;
  return l + 1;
}
int main() {
  int n, l, r;
  while (~scanf("%d%d%d", &n, &l, &r)) {
    int cnt[5] = {0}, dp[2][5] = {0}, t = 0;
    for (int i = 0; i < 3; i++) {
      cnt[i] = cal(r, i) - cal(l - 1, i);
    }
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      t ^= 1;
      for (int j = 0; j < 3; j++) {
        dp[t][j] = 0;
        for (int k = 0; k < 3; k++) {
          dp[t][j] += 1ll * cnt[k] * dp[t ^ 1][(j - k + 3) % 3] % mod;
          if (dp[t][j] >= mod) dp[t][j] -= mod;
        }
      }
    }
    printf("%d\n", dp[t][0]);
  }
  return 0;
}
