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 long long inf = 1e18;
long long l, r;
long long cnt[3];
long long dp[200005][3];
long long count(long long low, long long high, long long rem) {
  while (low <= high and low % 3 != rem) low += 1;
  while (high >= low and high % 3 != rem) high -= 1;
  if (low > high) return 0;
  return (high - low) / 3 + 1;
}
long long solve(long long n, long long sm) {
  if (n < 0) {
    return sm % 3 == 0 ? 1 : 0;
  }
  if (dp[n][sm] != -1) return dp[n][sm];
  long long res1 = 0, res2 = 0, res3 = 0;
  if (cnt[0] > 0) {
    res1 = (solve(n - 1, (sm + 0) % 3) * (cnt[0])) % 1000000007;
  }
  if (cnt[1] > 0) {
    res2 = (solve(n - 1, (sm + 1) % 3) * (cnt[1])) % 1000000007;
  }
  if (cnt[2] > 0) {
    res3 = (solve(n - 1, (sm + 2) % 3) * (cnt[2])) % 1000000007;
  }
  return dp[n][sm] = (res1 + res2 + res3) % 1000000007;
}
int main() {
  long long n;
  cin >> n >> l >> r;
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j < 3; j++) dp[i][j] = -1;
  for (long long i = 0; i < 3; i++) {
    cnt[i] = count(l, r, i);
  }
  long long res = solve(n - 1, 0);
  cout << res;
}
