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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
const ll oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
const ll N = 2e5 + 10;
ll dp[N][4];
ll rem0 = 0, rem1 = 0, rem2 = 0;
ll n, l, r;
ll fun(ll i, ll sum) {
  if (i == n) return sum == 0;
  auto& res = dp[i][sum];
  if (res != -1) return res;
  ll ans = 0;
  ans = (ans + (rem0 * fun(i + 1, sum + 0)) % MOD) % MOD;
  ans = (ans + (rem1 * fun(i + 1, (sum + 1) % 3)) % MOD) % MOD;
  ans = (ans + (rem2 * fun(i + 1, (sum + 2) % 3)) % MOD) % MOD;
  return res = ans;
}
signed main() {
  memset(dp, -1, sizeof dp);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> l >> r;
  while (l < min(3LL, r)) {
    if (l == 1) rem1++;
    if (l == 2) rem2++;
    l++;
  }
  if (r == 2) rem2++;
  if (r == 1) rem1++;
  rem0 += (r / 3 - (l - 1) / 3);
  l--;
  r--;
  rem1 += (r / 3 - (l - 1) / 3);
  l--;
  r--;
  rem2 += (r / 3 - (l - 1) / 3);
  rem0 = max(0LL, rem0);
  rem1 = max(0LL, rem1);
  rem2 = max(0LL, rem2);
  cout << fun(0, 0) << "\n";
  return 0;
}
