Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

C. DNA Alignment

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputVasya became interested in bioinformatics. He's going to write an article about similar cyclic DNA sequences, so he invented a new method for determining the similarity of cyclic sequences.

Let's assume that strings *s* and *t* have the same length *n*, then the function *h*(*s*, *t*) is defined as the number of positions in which the respective symbols of *s* and *t* are the same. Function *h*(*s*, *t*) can be used to define the function of Vasya distance ρ(*s*, *t*):

Vasya found a string *s* of length *n* on the Internet. Now he wants to count how many strings *t* there are such that the Vasya distance from the string *s* attains maximum possible value. Formally speaking, *t* must satisfy the equation: .

Vasya could not try all possible strings to find an answer, so he needs your help. As the answer may be very large, count the number of such strings modulo 10^{9} + 7.

Input

The first line of the input contains a single integer *n* (1 ≤ *n* ≤ 10^{5}).

The second line of the input contains a single string of length *n*, consisting of characters "ACGT".

Output

Print a single number — the answer modulo 10^{9} + 7.

Examples

Input

1

C

Output

1

Input

2

AG

Output

4

Input

3

TTT

Output

1

Note

Please note that if for two distinct strings *t*_{1} and *t*_{2} values ρ(*s*, *t*_{1}) и ρ(*s*, *t*_{2}) are maximum among all possible *t*, then both strings must be taken into account in the answer even if one of them can be obtained by a circular shift of another one.

In the first sample, there is ρ("*C*", "*C*") = 1, for the remaining strings *t* of length 1 the value of ρ(*s*, *t*) is 0.

In the second sample, ρ("*AG*", "*AG*") = ρ("*AG*", "*GA*") = ρ("*AG*", "*AA*") = ρ("*AG*", "*GG*") = 4.

In the third sample, ρ("*TTT*", "*TTT*") = 27

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Jun/23/2017 21:55:27 (c4).

Desktop version, switch to mobile version.

User lists

Name |
---|