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].
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Solution {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());

		int n = Integer.parseInt(st.nextToken());
		int l = Integer.parseInt(st.nextToken());
		int r = Integer.parseInt(st.nextToken());

		long[][] dp = new long[n + 1][3];

		int no = r - l + 1;

		long cn0 = 0;
		long cn1 = 0;
		long cn2 = 0;

		if (l % 3 == 0) {
			cn0 = (no - 1) / 3 + 1;
			if (no % 3 == 1) {
				cn1 = cn0 - 1;
				cn2 = cn0 - 1;
			} else if (no % 3 == 2) {
				cn1 = cn0;
				cn2 = cn0 - 1;
			} else {
				cn1 = cn0;
				cn2 = cn0;
			}
		} else if (l % 3 == 1) {
			cn1 = (no - 1) / 3 + 1;
			if (no % 3 == 1) {
				cn2 = cn1 - 1;
				cn0 = cn1 - 1;
			} else if (no % 3 == 2) {
				cn2 = cn1;
				cn0 = cn1 - 1;
			} else {
				cn2 = cn1;
				cn0 = cn1;
			}
		} else {
			cn2 = (no - 1) / 3 + 1;
			if (no % 3 == 1) {
				cn0 = cn2 - 1;
				cn1 = cn2 - 1;
			} else if (no % 3 == 2) {
				cn0 = cn2;
				cn1 = cn2 - 1;
			} else {
				cn0 = cn2;
				cn1 = cn2;
			}
		}

		int MOD = 1_000_000_007;

		dp[1][0] = cn0;
		dp[1][1] = cn1;
		dp[1][2] = cn2;

		for (int i = 2; i < dp.length; i++) {
			dp[i][0] = (dp[i][0] + (dp[i - 1][0] * cn0) % MOD) % MOD;
			dp[i][0] = (dp[i][0] + (dp[i - 1][1] * cn2) % MOD) % MOD;
			dp[i][0] = (dp[i][0] + (dp[i - 1][2] * cn1) % MOD) % MOD;

			dp[i][1] = (dp[i][1] + (dp[i - 1][0] * cn1) % MOD) % MOD;
			dp[i][1] = (dp[i][1] + (dp[i - 1][1] * cn0) % MOD) % MOD;
			dp[i][1] = (dp[i][1] + (dp[i - 1][2] * cn2) % MOD) % MOD;

			dp[i][2] = (dp[i][2] + (dp[i - 1][0] * cn2) % MOD) % MOD;
			dp[i][2] = (dp[i][2] + (dp[i - 1][1] * cn1) % MOD) % MOD;
			dp[i][2] = (dp[i][2] + (dp[i - 1][2] * cn0) % MOD) % MOD;
		}

		System.out.println(dp[n][0]);

	}

}