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 const mod = 1000000007;
long long const md = 998244353;
long long mypowr(long long a, long long b) {
  long long res = 1;
  if (b < 0) b = 0;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % md;
    a = a * a % md;
  }
  return res;
}
long long mypow(long long a, long long b) {
  long long res = 1;
  if (b < 0) b = 0;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
ifstream in("input.txt");
ofstream out("output.txt");
int u[5];
long long dp[200005][5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  long long n, l, r;
  cin >> n >> l >> r;
  if (r - l <= 100) {
    for (int i = l; i <= r; ++i) {
      ++u[i % 3];
    }
  } else {
    while (l % 3 != 0) {
      ++u[l % 3];
      ++l;
    }
    while (r % 3 != 0) {
      ++u[r % 3];
      --r;
    }
    u[0] += (r - l) / 3 + 1;
    u[1] += (r - l) / 3;
    u[2] += (r - l) / 3;
  }
  dp[1][0] = u[0];
  dp[1][1] = u[1];
  dp[1][2] = u[2];
  for (int i = 2; i <= n; ++i) {
    dp[i][0] = dp[i - 1][0] * u[0];
    dp[i][0] %= mod;
    dp[i][0] += dp[i - 1][1] * u[2];
    dp[i][0] %= mod;
    dp[i][0] += dp[i - 1][2] * u[1];
    dp[i][0] %= mod;
    dp[i][1] = dp[i - 1][1] * u[0];
    dp[i][1] %= mod;
    dp[i][1] += dp[i - 1][0] * u[1];
    dp[i][1] %= mod;
    dp[i][1] += dp[i - 1][2] * u[2];
    dp[i][1] %= mod;
    dp[i][2] = dp[i - 1][0] * u[2];
    dp[i][2] %= mod;
    dp[i][2] += dp[i - 1][2] * u[0];
    dp[i][2] %= mod;
    dp[i][2] += dp[i - 1][1] * u[1];
    dp[i][2] %= mod;
  }
  cout << dp[n][0] << '\n';
  return 0;
}
