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 int mod_ = 1000000007;
vector<long long int> dp1;
vector<long long int> dp2;
vector<long long int> dp0;
int main() {
  int n, l, r;
  cin >> n >> l >> r;
  if (n == 0) return 0;
  if (l == r) {
    if (l % 3 == 0 || n % 3 == 0)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  dp0.resize(n + 1, -1);
  dp1.resize(n + 1, -1);
  dp2.resize(n + 1, -1);
  dp0[0] = 0;
  dp1[0] = 0;
  dp2[0] = 0;
  long long int st1, st2, st0, end1, end2, end0;
  long long int mod1 = 0, mod2 = 0, mod0 = 0;
  if (l % 3 == 0) {
    st0 = l;
    st1 = l + 1;
    st2 = l + 2;
  } else if (l % 3 == 1) {
    st0 = l + 2;
    st1 = l;
    st2 = l + 1;
  } else {
    st0 = l + 1;
    st1 = l + 2;
    st2 = l;
  }
  if (r % 3 == 0) {
    end0 = r;
    end1 = r - 2;
    end2 = r - 1;
  } else if (r % 3 == 1) {
    end0 = r - 1;
    end1 = r;
    end2 = r - 2;
  } else {
    end0 = r - 2;
    end1 = r - 1;
    end2 = r;
  }
  if (end0 < st0)
    mod0 = 0;
  else
    mod0 = (end0 - st0) / 3 + 1;
  if (end1 < st1)
    mod1 = 0;
  else
    mod1 = (end1 - st1) / 3 + 1;
  if (end2 < st2)
    mod2 = 0;
  else
    mod2 = (end2 - st2) / 3 + 1;
  dp0[1] = mod0;
  dp1[1] = mod1;
  dp2[1] = mod2;
  for (int i = 2; i <= n; i++) {
    dp0[i] = (((dp0[i - 1] * mod0) % mod_ + (dp1[i - 1] * mod2) % mod_) % mod_ +
              (dp2[i - 1] * mod1) % mod_) %
             mod_;
    dp1[i] = (((dp0[i - 1] * mod1) % mod_ + (dp1[i - 1] * mod0) % mod_) % mod_ +
              (dp2[i - 1] * mod2) % mod_) %
             mod_;
    dp2[i] = (((dp0[i - 1] * mod2) % mod_ + (dp1[i - 1] * mod1) % mod_) % mod_ +
              (dp2[i - 1] * mod0) % mod_) %
             mod_;
  }
  cout << dp0[n];
  return 0;
}
