Ayoub and Lost Array

C. Ayoub and Lost Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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].

#include<bits/stdc++.h>
#define REP(i, a, b) for(int i = (a); i <= (b); ++ i)
#define REP(j, a, b) for(int j = (a); j <= (b); ++ j)
#define PER(i, a, b) for(int i = (a); i >= (b); -- i)
using namespace std;
const int maxn=2e5+5;
const int mod=1e9+7;
template <class T>
inline void rd(T &ret){
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9'){
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}
int n,l,r;
long long dp[maxn][5];
int main()
{
    rd(n),rd(l),rd(r);
    int u=r/3-(l-1)/3;
    int v=(r+2)/3-(l-1+2)/3;
    int w=(r+1)/3-(l-1+1)/3;
    dp[1][0]=u,dp[1][1]=v,dp[1][2]=w;
    REP(i,2,n){
        dp[i][0]=(dp[i-1][2]*v%mod+dp[i-1][1]*w%mod+dp[i-1][0]*u%mod)%mod;
        dp[i][1]=(dp[i-1][0]*v%mod+dp[i-1][1]*u%mod+dp[i-1][2]*w%mod)%mod;
        dp[i][2]=(dp[i-1][0]*w%mod+dp[i-1][1]*v%mod+dp[i-1][2]*u%mod)%mod;
    }
    cout<<dp[n][0]<<endl;
}