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;
template <class T>
inline T bigMod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline string int2String(T a) {
  ostringstream str;
  str << a;
  return str.str();
}
template <class T>
inline T string2int(string a) {
  stringstream str(a);
  T b;
  str >> b;
  return b;
}
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1, -2, -2, 2, 2, -1, -1, 1, 1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1, -1, 1, -1, 1, -2, 2, -2, 2};
long long DP[200005][3], n, l, r;
long long inside(long long l, long long r, long long i) {
  while (l % 3 != i) l++;
  while (r >= l && r % 3 != i) r--;
  if (l > r) return 0;
  return (r - l + 3LL) / 3LL;
}
long long solve(int pos, int m) {
  if (pos == n) return (m == 0);
  long long &ret = DP[pos][m];
  if (~ret) return ret;
  ret = 0;
  for (long long i = 0; i < 3; i++) {
    ret += inside(l, r, i) * solve(pos + 1, (i + m) % 3);
    ret %= 1000000007;
  }
  return ret;
}
int main() {
  memset(DP, -1, sizeof(DP));
  scanf("%lld %lld %lld", &n, &l, &r);
  printf("%lld\n", solve(0, 0));
  return 0;
}
