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;
char s[200010];
int fact[200001];
long long d[200001][3];
int ridput(int x, int y) {
  int sol = 1;
  while (y) {
    if (y % 2) sol = ((long long)sol * x) % 1000000007;
    x = ((long long)x * x) % 1000000007;
    y /= 2;
  }
  return sol;
}
int comb(int n, int k) {
  return ((long long)fact[n] *
          ridput(((long long)fact[k] * fact[n - k]) % 1000000007,
                 1000000007 - 2)) %
         1000000007;
}
int main() {
  int n, l, r, i, lf, rg, r1, r2, r0, sol;
  scanf("%d%d%d", &n, &l, &r);
  fact[0] = 1;
  for (i = 1; i <= n; i++) fact[i] = ((long long)fact[i - 1] * i) % 1000000007;
  lf = l / 3;
  rg = r / 3;
  r1 = r2 = r0 = rg - lf + 1;
  if (l % 3 == 1)
    r0--;
  else if (l % 3 == 2) {
    r0--;
    r1--;
  }
  if (r % 3 == 1) {
    r2--;
  } else if (r % 3 == 0) {
    r2--;
    r1--;
  }
  if (r0) d[1][0] = r0;
  if (r1) d[1][1] = r1;
  if (r2) d[1][2] = r2;
  for (i = 2; i <= n; i++) {
    if (r0) {
      d[i][0] = (d[i][0] + (long long)d[i - 1][0] * r0) % 1000000007;
      d[i][1] = (d[i][1] + (long long)d[i - 1][1] * r0) % 1000000007;
      d[i][2] = (d[i][2] + (long long)d[i - 1][2] * r0) % 1000000007;
    }
    if (r1) {
      d[i][0] = (d[i][0] + (long long)d[i - 1][2] * r1) % 1000000007;
      d[i][1] = (d[i][1] + (long long)d[i - 1][0] * r1) % 1000000007;
      d[i][2] = (d[i][2] + (long long)d[i - 1][1] * r1) % 1000000007;
    }
    if (r2) {
      d[i][0] = (d[i][0] + (long long)d[i - 1][1] * r2) % 1000000007;
      d[i][1] = (d[i][1] + (long long)d[i - 1][2] * r2) % 1000000007;
      d[i][2] = (d[i][2] + (long long)d[i - 1][0] * r2) % 1000000007;
    }
  }
  printf("%lld", d[n][0]);
  return 0;
}
