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;
void io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isperfect(long long n) {
  long long y = sqrt(n);
  if (n % y == 0 && y * y == n) return true;
  return false;
}
bool comp(long long a, long long b) { return a > b; }
long long powi(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    b = b >> 1;
    a = a * a;
  }
  return ans;
}
bool isprime(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
bool prime[10000000];
void seive() {
  prime[1] = true;
  for (int i = 2; i * i < 10000000; i++) {
    if (prime[i]) continue;
    for (int j = i * i; j < 10000000; j += i) {
      prime[j] = true;
    }
  }
}
int kmp(string t) {
  int n = t.length();
  int kmp[n];
  kmp[0] = 0;
  int i = 0, r = 1;
  while (r < n) {
    if (t[r] == t[i]) {
      kmp[r] = i + 1;
      i++;
      r++;
    } else {
      if (i == 0) {
        kmp[r] = 0;
        r++;
      } else
        i = kmp[i - 1];
    }
  }
  return kmp[n - 1];
}
int main() {
  io();
  int n, l, r;
  cin >> n >> l >> r;
  long long dp[n + 1][3];
  long long mod0, mod1 = r / 3 - (l - 1) / 3, mod2 = r / 3 - (l - 1) / 3;
  mod0 = (r) / 3 - (l - 1) / 3;
  if (r % 3 == 1) {
    mod1++;
  }
  if (r % 3 == 2) {
    mod1++, mod2++;
  }
  if ((l - 1) % 3 == 1) mod1--;
  if ((l - 1) % 3 == 2) {
    mod1--;
    mod2--;
  }
  dp[1][0] = mod0, dp[1][1] = mod1, dp[1][2] = mod2;
  for (int i = 2; i < n + 1; i++) {
    dp[i][0] =
        (dp[i - 1][0] * mod0 + dp[i - 1][1] * mod2 + dp[i - 1][2] * mod1) %
        1000000007;
    dp[i][1] =
        (dp[i - 1][0] * mod1 + dp[i - 1][1] * mod0 + dp[i - 1][2] * mod2) %
        1000000007;
    dp[i][2] =
        (dp[i - 1][0] * mod2 + dp[i - 1][1] * mod1 + dp[i - 1][2] * mod0) %
        1000000007;
  }
  cout << dp[n][0] % 1000000007;
}
