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 MAXN = 200000;
const long long mod = 1e9 + 7;
long long inv[MAXN + 5];
void init() {
  inv[1] = 1;
  for (int i = 2; i <= MAXN; i++) {
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  }
}
long long dp[MAXN + 5][4];
long long getAns(int n, int l, int r) {
  int cnt[4] = {0};
  if (r - l < 10) {
    for (int i = l; i <= r; i++) {
      cnt[i % 3]++;
    }
  } else {
    while (l % 3 != 0) {
      cnt[l % 3]++;
      l++;
    }
    while (r % 3 != 2) {
      cnt[r % 3]++;
      r--;
    }
    cnt[0] += (r - l + 1) / 3;
    cnt[1] += (r - l + 1) / 3;
    cnt[2] += (r - l + 1) / 3;
  }
  dp[0][0] = 1;
  dp[0][1] = 0;
  dp[0][2] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 3; j++) {
      dp[i][j] = 0;
      for (int k = 0; k < 3; k++) {
        dp[i][j] += dp[i - 1][(j - k + 3) % 3] * cnt[k];
        dp[i][j] %= mod;
      }
    }
  }
  return dp[n][0];
}
int main() {
  init();
  int n, l, r;
  while (cin >> n >> l >> r) {
    cout << getAns(n, l, r) << endl;
  }
}
