Ayoub and Lost Array【dp】

Ayoub had an array aa of integers of size nn and this array had two interesting properties:

• All the integers in the array were between ll and rr (inclusive).
• The sum of all the elements was divisible by 33.

Unfortunately, Ayoub has lost his array, but he remembers the size of the array nn and the numbers ll and rr, 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+7109+7 (i.e. the remainder when dividing by 109+7109+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 00.

Input

The first and only line contains three integers nn, ll and rr (1≤n≤2⋅105,1≤l≤r≤1091≤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+7109+7 the number of ways to restore the array.

Examples

input

Copy

2 1 3

output

Copy

3

input

Copy

3 2 2

output

Copy

1

input

Copy

9 9 99

output

Copy

711426616

Note

In the first example, the possible arrays are : [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].

 

 

题意：

从l到r中取n个数字组成一个数组，这个数组所有元素的和可以整除3，求最多的方案数。

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
ll mod=1e9+7;
ll dp[200005][4];
int main()
{
    ll n,l,r;
    while(cin>>n>>l>>r)
    {
        memset(dp,0,sizeof(dp));
        dp[1][0]=r/3-(l-1)/3;//l，r之间取余3mod0的
        dp[1][1]=(r+2)/3-(l+1)/3;//l，r之间取余3mod1的
        dp[1][2]=(r+1)/3-l/3;//l，r之间取余3mod2的
        for(int i=2; i<=n; i++)
        {
            dp[i][0]=(dp[i-1][0]*dp[1][0]%mod+dp[i-1][1]*dp[1][2]%mod+dp[i-1][2]*dp[1][1]%mod)%mod;///前i的几个数取余为0的数量=
            ///前i-1的几个数取余为0的数量  +  前i-1个数取余为1的数量乘以l到r中取余为2的数字的数量  +  前i-1个数的和取余为2的数量乘以l到r的数中取余为1的数量。


            dp[i][1]=(dp[i-1][0]*dp[1][1]%mod+dp[i-1][1]*dp[1][0]%mod+dp[i-1][2]*dp[1][2]%mod)%mod;//前i的几个数取余为1的数量
            dp[i][2]=(dp[i-1][0]*dp[1][2]%mod+dp[i-1][1]*dp[1][1]%mod+dp[i-1][2]*dp[1][0]%mod)%mod;//前i的几个数取余为2的数量
        }
        cout<<dp[n][0]<<endl;
    }
    return 0;
}