problem

Cryptography is all the rage at xryuseix's school. Xryuseix, who lives in a grid of cities, has come up with a new cryptography to decide where to meet.

The ciphertext consists of the $ N $ character string $ S $, and the $ S_i $ character determines the direction of movement from the current location. The direction of movement is as follows.


* A ~ M: Go north one square.
* N ~ Z: Go south one square.
* a ~ m: Go east one square.
* n ~ z: Go west one square.



By the way, xryuseix wanted to tell yryuseiy-chan where to meet for a date with a ciphertext, but he noticed that the ciphertext was redundant.

For example, suppose you have the ciphertext "ANA". It goes north by $ 1 $, south by $ 1 $, and then north by $ 1 $. This is equivalent to the ciphertext that goes north by $ 1 $. , "ANA" = "A", which can be simplified. Xryuseix wanted to simplify the ciphertext so that yryuseiy would not make a detour.

So you decided to write a program to simplify the ciphertext instead of xryuseix. Note that "simplify the ciphertext" means "the shortest ciphertext that goes to the same destination as the original ciphertext." To make. "



output

The length of the ciphertext after simplification on the $ 1 $ line, output the ciphertext on the $ 2 $ line. If there are multiple possible ciphertexts as an answer, any of them may be output. Also, each line Output a line break at the end of.

Example

Input

5
ANazA


Output

1
A
#include "bits/stdc++.h"
#define YES "YES"
#define NO "NO"
#define YESNO OUT(three(solve(),YES,NO))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(b);i>=(LL)(a);i--)
#define REP(i,b) FOR(i,zero,b)
#define rep REP
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,b,zero)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB push_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define zero 0LL
#define all ALL
#define pb emplace_back
#define eb pb
#define int long long
using namespace std;
template<typename T> inline void maximize(T &a, T b) { a = max(a, b); }
template<typename T> inline void minimize(T &a, T b) { a = min(a, b); }
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef int LL;
typedef double ld;
typedef int ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef tuple<int, int, int, int> tapu;
typedef vector<tapu> Vtapu;
typedef priority_queue<tapu, Vtapu, greater<tapu> > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
const int SIZE1 = 3e5 + 1000;
const int SIZE2 = 5010;
const int SIZE3 = 430;
const int SIZE = SIZE1;
const int MAPSIZE = 40;
const LL p = 7 + 1e9;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
typedef pair<ld, ut> pld;
 
int dx[]={0,0,1,-1};
int dy[]={-1,1,0,0};
signed main(){
  LL A,B,X;
  string s;
  LL N;
  cin >> N;
  cin >> s;
  LL x=0,y=0;
  char st[]={'A','N','a','n'};
  char ed[]={'M','Z','m','z'};
  REP(i,N){
    REP(j,4)
     if(st[j]<=s[i] and s[i]<=ed[j]){
      x+=dx[j];
      y+=dy[j];
     }
  }
//  cout << x << "" << y << endl;
  cout << abs(x)+abs(y) << endl;
  REP(j,4){
    FOR(i,1,SIZE) if((dx[j]*i== x and dx[j]) or (dy[j] and dy[j]*i==y)){
    //  cout << i << " " << j << endl;

      REP(k,i) cout <<st[j];
    }
  }
  cout << endl;
 	return 0;
}

