Codeforces 1105C： Ayoub and Lost Array（递推）

time limit per test: 1 second

memory limit per test: 256 megabytes

input: standard input

output: standard output

Ayoub had an array aaa of integers of size nnn and this array had two interesting properties:

• All the integers in the array were between lll and rrr (inclusive).
• The sum of all the elements was divisible by 333.

Unfortunately, Ayoub has lost his array, but he remembers the size of the array nnn and the numbers lll and rrr, so he asked you to find the number of ways to restore the array.

Since the answer could be very large, print it modulo 109+710^9+7109+7 (i.e. the remainder when dividing by 109+710^9+7109+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 000.

Input

The first and only line contains three integers nnn, lll and rrr (1≤n≤2⋅105,1≤l≤r≤109)(1≤n≤2⋅10^5,1≤l≤r≤10^9)(1≤n≤2⋅105,1≤l≤r≤109) — the size of the lost array and the range of numbers in the array.

Output

Print the remainder when dividing by 109+710^9+7109+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][1,2],[2,1],[3,3][1,2],[2,1],[3,3].

In the second example, the only possible array is [2,2,2][2,2,2][2,2,2].

题意

给出三个整数n,l,rn,l,rn,l,r，要求在[l,r][l,r][l,r]之间的数组成的长度为nnn的序列的和能够整除333

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#include <time.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
#define bug cout<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
ll dp[3][maxn];
int n, l, r, t ;
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	cin>>n>>l>>r;
	int num0,num1,num2;//余数为i的个数
	num0=r/3-(l-1)/3;
	num1=r/3+(r%3>=2)-(l-1)/3-((l-1)%3>=2);
	num2=r-l+1-num0-num1;
	dp[0][1]=num0;
	dp[1][1]=num1;
	dp[2][1]=num2;
	for(int i=2;i<=n;i++)
	{
		dp[0][i]=((dp[0][i-1]%mod*(num0%mod))%mod+(dp[1][i-1]%mod*num2%mod)%mod+(dp[2][i-1]%mod*num1%mod)%mod)%mod;
		dp[1][i]=((dp[0][i-1]%mod*(num1%mod))%mod+(dp[1][i-1]%mod*num0%mod)%mod+(dp[2][i-1]%mod*num2%mod)%mod)%mod;
		dp[2][i]=((dp[0][i-1]%mod*(num2%mod))%mod+(dp[1][i-1]%mod*num1%mod)%mod+(dp[2][i-1]%mod*num0%mod)%mod)%mod;
	}
	cout<<dp[0][n]%mod<<endl;
	return 0;
}