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.util.*;
import java.io.*;
import java.util.concurrent.TimeUnit;

public class ayoubLost {
	static int L, R, N;
	static final int MOD = 1000000007;
	
	static long[][] storage;
	
	static int zero, one, two;
	public static void main (String [] args) throws Exception {
	    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	    StringTokenizer st = new StringTokenizer(f.readLine());
	    
	    N = Integer.parseInt(st.nextToken());
	    L = Integer.parseInt(st.nextToken());
	    R = Integer.parseInt(st.nextToken());
	    
	    for(; L <= R; L++) {
	    	if(L % 3 == 0)
	    		zero++;
	    	else if(L % 3 == 1)
	    		one++;
	    	else
	    		two++;
	    	if(L % 3 == 0) {
	    		break;
	    	}
	    }
	    for(; L <= R; R--) {
	    	if(R % 3 == 0)
	    		break;
	    	if(R % 3 == 0)
	    		zero++;
	    	else if(R % 3 == 1)
	    		one++;
	    	else
	    		two++;
	    }
	    int temp = (R-L)/3;
	    if(temp > 0) {
	    	zero += temp;
	    	one += temp;
	    	two += temp;
	    }
	    zero %= MOD;
	    one %= MOD;
	    two %=MOD;
	    //System.out.println(L + " " + R);
	    //System.out.println(zero + " " + one + " " + two);
	    
	    storage = new long[N][3];
	    out.println(dp(0, 0));
	    
	    //for(long[] each: storage)
	    //	System.out.println(Arrays.toString(each));
	    
	    out.close();
	    f.close();
	}
	
	public static long dp(int index, int curMod) {
		if(index == N) {
			if(curMod % 3 == 0)
				return 1;
			return 0;
		}
		if(storage[index][curMod] != 0) {
			return storage[index][curMod];
		}
		long sum = 0;
		sum += (dp(index+1, (curMod) % 3) % MOD * zero) % MOD;
		sum += (dp(index+1, (curMod+1) % 3) % MOD * one) % MOD;
		sum += (dp(index+1, (curMod+2) % 3) % MOD * two) % MOD;

		storage[index][curMod] = sum % MOD;
		return storage[index][curMod];
	}
	
}