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;
long long DP[200001][3];
int main() {
  long long mod = 1e9 + 7;
  int n, l, r;
  scanf("%d %d %d", &n, &l, &r);
  int S0 = 0, S1 = 0, S2 = 0;
  if (l % 3 == 1) {
    S1++;
    S2++;
    l += 2;
  }
  if (l % 3 == 2) {
    S2++;
    l++;
  }
  if (r % 3 == 0) {
    S0++;
    r--;
  }
  if (r % 3 == 1) {
    S1++;
    S0++;
    r -= 2;
  }
  S0 += (r - l + 1) / 3;
  S1 += (r - l + 1) / 3;
  S2 += (r - l + 1) / 3;
  DP[1][0] = S0;
  DP[1][1] = S1;
  DP[1][2] = S2;
  for (int i = 2; i <= n; i++) {
    DP[i][0] = ((DP[i - 1][0] * S0) % mod + (DP[i - 1][1] * S2) % mod +
                (DP[i - 1][2] * S1) % mod) %
               mod;
    DP[i][1] = ((DP[i - 1][0] * S1) % mod + (DP[i - 1][1] * S0) % mod +
                (DP[i - 1][2] * S2) % mod) %
               mod;
    DP[i][2] = ((DP[i - 1][0] * S2) % mod + (DP[i - 1][1] * S1) % mod +
                (DP[i - 1][2] * S0) % mod) %
               mod;
  }
  printf("%lld\n", DP[n][0]);
  return 0;
}
