You are given a string s consisting of n characters. These characters are among the first k lowercase letters of the Latin alphabet. You have to perform n operations with the string.

During the i-th operation, you take the character that initially occupied the i-th position, and perform one of the following actions with it:

  * swap it with the previous character in the string (if it exists). This operation is represented as L; 
  * swap it with the next character in the string (if it exists). This operation is represented as R; 
  * cyclically change it to the previous character in the alphabet (b becomes a, c becomes b, and so on; a becomes the k-th letter of the Latin alphabet). This operation is represented as D; 
  * cyclically change it to the next character in the alphabet (a becomes b, b becomes c, and so on; the k-th letter of the Latin alphabet becomes a). This operation is represented as U; 
  * do nothing. This operation is represented as 0. 



For example, suppose the initial string is test, k = 20, and the sequence of operations is URLD. Then the string is transformed as follows:

  1. the first operation is U, so we change the underlined letter in test to the next one in the first 20 Latin letters, which is a. The string is now aest; 
  2. the second operation is R, so we swap the underlined letter with the next one in the string aest. The string is now aset; 
  3. the third operation is L, so we swap the underlined letter with the previous one in the string aset (note that this is now the 2-nd character of the string, but it was initially the 3-rd one, so the 3-rd operation is performed to it). The resulting string is saet; 
  4. the fourth operation is D, so we change the underlined letter in saet to the previous one in the first 20 Latin letters, which is s. The string is now saes. 



The result of performing the sequence of operations is saes.

Given the string s and the value of k, find the lexicographically smallest string that can be obtained after applying a sequence of operations to s.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line contains two integers n and k (1 ≤ n ≤ 500; 2 ≤ k ≤ 26). 

The second line contains a string s consisting of n characters. Each character is one of the k first letters of the Latin alphabet (in lower case).

Output

For each test case, print one line containing the lexicographically smallest string that can be obtained from s using one sequence of operations.

Example

Input


6
4 2
bbab
7 5
cceddda
6 5
ecdaed
7 4
dcdbdaa
8 3
ccabbaca
5 7
eabba


Output


aaaa
baccacd
aabdac
aabacad
aaaaaaaa
abadb

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;
typedef double db;
typedef string str;

typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<str> vstr;

#define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in)
#define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in)
#define REP(i,b) FOR(i,0,b,1)
#define REPD(i,b) FORD(i,b,0,1)
#define pb push_back 
#define mp make_pair
#define ff first
#define ss second
#define all(x) begin(x), end(x)
#define rsz resize 
#define MANY_TESTS int tcase; cin >> tcase; while(tcase--)
	
const double EPS = 1e-9;
const int MOD = 1e9+7; // 998244353;
const ll INFF = 1e18;
const int INF = 1e9;
const ld PI = acos((ld)-1);
const vi dy = {1, 0, -1, 0, -1, 1, 1, -1};
const vi dx = {0, 1, 0, -1, -1, 1, -1, 1};

#ifdef DEBUG
#define DBG if(1)
#else
#define DBG if(0)
#endif

#define dbg(x) cout << "(" << #x << " : " << x << ")" << endl;
// ostreams
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;}
template <class T>
ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template<class T>
ostream& operator<<(ostream& out, const vector<T> &v){ out << "[";  REP(i, v.size()) out << v[i] << ", ";  out << "]"; return out;}
// istreams
template<class T>
istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; }
template<class T, class U>
istream& operator>>(istream& in, pair<T, U> &p){ in >> p.ff >> p.ss; return in; }
//searches
template<typename T, typename U>
T bsl(T lo, T hi, U f){ hi++; T mid; while(lo < hi){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid+1; } return lo; }
template<typename U>
double bsld(double lo, double hi, U f, double p = 1e-9){ int r = 3 + (int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid; } return (lo + hi)/2; }
template<typename T, typename U>
T bsh(T lo, T hi, U f){ lo--; T mid; while(lo < hi){ mid = (lo + hi + 1)/2; f(mid) ? lo = mid : hi = mid-1; } return lo; }
template<typename U>
double bshd(double lo, double hi, U f, double p = 1e-9){ int r = 3+(int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? lo = mid : hi = mid; } return (lo + hi)/2; }
// some more utility functions
template<typename T>
pair<T, int> get_min(vector<T> &v){ typename vector<T> :: iterator it = min_element(v.begin(), v.end()); return mp(*it, it - v.begin());}
template<typename T>
pair<T, int> get_max(vector<T> &v){ typename vector<T> :: iterator it = max_element(v.begin(), v.end()); return mp(*it, it - v.begin());}        
template<typename T> bool ckmin(T& a, const T& b){return b < a ? a = b , true : false;}
template<typename T> bool ckmax(T& a, const T& b){return b > a ? a = b, true : false;}

    
int main(){
 	ios_base::sync_with_stdio(false);
 	cin.tie(NULL); cout.tie(NULL);
    MANY_TESTS{
        int n, k; cin >> n >> k;
        string s; cin >> s;
        vector<string> dp(n+1, "-1");
        dp[0] = "";
        auto rotup = [&](char x){
            int i = x - 'a';
            i++; if(i >= k) i-=k;
            return (char)(i + 'a');
        };
        auto rotdw = [&](char x){
            int i = x - 'a';
            i--; if(i < 0) i+= k;
            return (char)(i + 'a');
        };
        auto min_s = [](string a, string b){
            if(a == "-1") return b;
            else if(b == "-1") return a;
            else return min(a, b);
        };
        REP(i, n){
            // up
            dp[i+1] = min_s(dp[i+1], dp[i] + rotup(s[i]));
            // down
            dp[i+1] = min_s(dp[i+1], dp[i] + rotdw(s[i]));
            // left
            if(i > 0){
                dp[i+1] = min_s(dp[i+1], dp[i].substr(0, i-1) + s[i] + dp[i][i-1]);
            }
            // RU, RD, Rl
            if(i < n - 1){
                dp[i+2] = min_s(dp[i+2], dp[i] + rotup(s[i+1]) + s[i]);
                dp[i+2] = min_s(dp[i+2], dp[i] + rotdw(s[i+1]) + s[i]);
                dp[i+2] = min_s(dp[i+2], dp[i] + s[i] + s[i+1]);
                if(i > 0)
                    dp[i+2] = min_s(dp[i+2], dp[i].substr(0, i-1) + s[i+1] + dp[i][i-1] + s[i]);
            }
        }
        cout << dp[n] << "\n";
    }
	
    return 0;
}
