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;
const int inf = 0x3f3f3f3f;
const long long INF = 1e18;
const int maxn = 2e5 + 7;
const long long mod = 1e9 + 7;
inline long long read() {
  long long x = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
void out(long long x) {
  int stackk[20];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  int top = 0;
  while (x) stackk[++top] = x % 10, x /= 10;
  while (top) putchar(stackk[top--] + '0');
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long n, r, l, dp[maxn][6], num[6];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> l >> r;
  num[0] = r / 3 - (l - 1) / 3;
  num[1] = (r + 2) / 3 - (l + 1) / 3;
  num[2] = (r + 1) / 3 - l / 3;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 2; j++) {
      for (int k = 0; k <= 2; k++) {
        int t = (j - k + 3) % 3;
        dp[i][j] = (dp[i][j] + dp[i - 1][t] * num[k]) % mod;
      }
    }
  }
  cout << dp[n][0];
  return 0;
}
