Codeforce-CodeCraft-20 (Div. 2)-B. String Modification (找规律+模拟）

Vasya has a string s of length n. He decides to make the following modification to the string:

Pick an integer k, (1≤k≤n).

For i from 1 to n−k+1, reverse the substring s[i:i+k−1] of s. For example, if string s is qwer and k=2, below is the series of transformations the string goes through:

qwer (original string)

wqer (after reversing the first substring of length 2)

weqr (after reversing the second substring of length 2)

werq (after reversing the last substring of length 2)

Hence, the resulting string after modifying s with k=2 is werq.

Vasya wants to choose a k such that the string obtained after the above-mentioned modification is lexicographically smallest possible among all choices of k. Among all such k, he wants to choose the smallest one. Since he is busy attending Felicity 2020, he asks for your help.

A string a is lexicographically smaller than a string b if and only if one of the following holds:

a is a prefix of b, but a≠b;

in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.

Input

Each test contains multiple test cases.

The first line contains the number of test cases t (1≤t≤5000). The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤5000) — the length of the string s.

The second line of each test case contains the string s of n lowercase latin letters.

It is guaranteed that the sum of n over all test cases does not exceed 5000.

Output

For each testcase output two lines:

In the first line output the lexicographically smallest string s′ achievable after the above-mentioned modification.

In the second line output the appropriate value of k (1≤k≤n) that you chose for performing the modification. If there are multiple values of k that give the lexicographically smallest string, output the smallest value of k among them.

Example

inputCopy

6

4

abab

6

qwerty

5

aaaaa

6

alaska

9

lfpbavjsm

1

p

outputCopy

abab

1

ertyqw

3

aaaaa

1

aksala

6

avjsmbpfl

5

p

1

Note

In the first testcase of the first sample, the string modification results for the sample abab are as follows :

for k=1 : abab

for k=2 : baba

for k=3 : abab

for k=4 : baba

The lexicographically smallest string achievable through modification is abab for k=1 and 3. Smallest value of k needed to achieve is hence 1.

//规律从后往前数，分奇偶，然后要么是直接放后面，要么是导致，直接写string模拟这过程完事了。

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    t f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}

#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 200005
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
int n;
string c;

int main()
{
    int t, f;
    read(t);
    while (t--)
    {
        read(n);
        cin >> c;
        f=0;
        string ans = c, s;
        for (int i = 1; i < n; i++)
        {
            if ((n - i) % 2 == 0)
                s = c.substr(i, n - i) + c.substr(0, i );
            else
            {
                string tem = c.substr(0, i );
                 reverse(tem.begin(), tem.end());
                s = c.substr(i, n - i) +tem;
            }
            if (ans > s)
            {
                f = i;
                ans = s;
            }
        }
        cout << ans << endl;

        wi(f + 1), P;
    }
}