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 N = 2e5 + 5;
const int M = 1e9 + 7;
void quickSort(int a[], int b[], int l, int r) {
  int p = a[(l + r) / 2];
  int i = l, j = r;
  while (i < j) {
    while (a[i] < p) {
      i++;
    }
    while (a[j] > p) {
      j--;
    }
    if (i <= j) {
      int temp = a[i];
      a[i] = a[j];
      a[j] = temp;
      temp = b[i];
      b[i] = b[j];
      b[j] = temp;
      i++;
      j--;
    }
  }
  if (i < r) {
    quickSort(a, b, i, r);
  }
  if (l < j) {
    quickSort(a, b, l, j);
  }
}
long long n, l, r, dp[N][3];
long long zero, one, two;
void remainder() {
  long long x, y;
  x = r;
  y = l;
  while (x > l && x % 3 != 0) --x;
  while (y < r && y % 3 != 0) ++y;
  zero = (x - y) / 3 + 1;
  x = r;
  y = l;
  while (x > l && x % 3 != 1) --x;
  while (y < r && y % 3 != 1) ++y;
  one = (x - y) / 3 + 1;
  x = r;
  y = l;
  while (x > l && x % 3 != 2) --x;
  while (y < r && y % 3 != 2) ++y;
  two = (x - y) / 3 + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> l >> r;
  if (r - l > 3)
    remainder();
  else {
    for (int i = l; i <= r; i++)
      if (i % 3 == 0)
        ++zero;
      else if (i % 3 == 1)
        ++one;
      else
        ++two;
  }
  dp[1][0] = zero;
  dp[1][1] = one;
  dp[1][2] = two;
  for (int i = 2; i <= n; i++) {
    dp[i][0] = (dp[i - 1][0] * zero % M + dp[i - 1][1] * two % M +
                dp[i - 1][2] * one % M) %
               M;
    dp[i][1] = (dp[i - 1][0] * one % M + dp[i - 1][1] * zero % M +
                dp[i - 1][2] * two % M) %
               M;
    dp[i][2] = (dp[i - 1][0] * two % M + dp[i - 1][1] * one % M +
                dp[i - 1][2] * zero % M) %
               M;
  }
  cout << dp[n][0];
}
