Problem: Boboniu defines BN-string as a string s of characters 'B' and 'N'.

You can perform the following operations on the BN-string s:

  * Remove a character of s. 
  * Remove a substring "BN" or "NB" of s. 
  * Add a character 'B' or 'N' to the end of s. 
  * Add a string "BN" or "NB" to the end of s. 



Note that a string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Boboniu thinks that BN-strings s and t are similar if and only if:

  * |s|=|t|. 
  * There exists a permutation p_1, p_2, …, p_{|s|} such that for all i (1≤ i≤ |s|), s_{p_i}=t_i. 



Boboniu also defines dist(s,t), the distance between s and t, as the minimum number of operations that makes s similar to t.

Now Boboniu gives you n non-empty BN-strings s_1,s_2,…, s_n and asks you to find a non-empty BN-string t such that the maximum distance to string s is minimized, i.e. you need to minimize max_{i=1}^n dist(s_i,t).

Input

The first line contains a single integer n (1≤ n≤ 3⋅ 10^5).

Each of the next n lines contains a string s_i (1≤ |s_i| ≤ 5⋅ 10^5). It is guaranteed that s_i only contains 'B' and 'N'. The sum of |s_i| does not exceed 5⋅ 10^5.

Output

In the first line, print the minimum max_{i=1}^n dist(s_i,t).

In the second line, print the suitable t.

If there are several possible t's, you can print any.

Examples

Input


3
B
N
BN


Output


1
BN


Input


10
N
BBBBBB
BNNNBBNBB
NNNNBNBNNBNNNBBN
NBNBN
NNNNNN
BNBNBNBBBBNNNNBBBBNNBBNBNBBNBBBBBBBB
NNNNBN
NBBBBBBBB
NNNNNN


Output


12
BBBBBBBBBBBBNNNNNNNNNNNN


Input


8
NNN
NNN
BBNNBBBN
NNNBNN
B
NNN
NNNNBNN
NNNNNNNNNNNNNNNBNNNNNNNBNB


Output


12
BBBBNNNNNNNNNNNN


Input


3
BNNNBNNNNBNBBNNNBBNNNNBBBBNNBBBBBBNBBBBBNBBBNNBBBNBNBBBN
BBBNBBBBNNNNNBBNBBBNNNBB
BBBBBBBBBBBBBBNBBBBBNBBBBBNBBBBNB


Output


12
BBBBBBBBBBBBBBBBBBBBBBBBBBNNNNNNNNNNNN

Note

In the first example dist(B,BN)=dist(N,BN)=1, dist(BN,BN)=0. So the maximum distance is 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long n, ma = 5e5, fq[2], d0[2], d1[2], d2[2], sq[2], inf = 1e18;
int main() {
  long long i, j, ii, u, k, ln, lb, rb, ub, p, z = inf;
  string s;
  scanf("%lld", &n);
  for (ii = 0; ii < 2; ii++) {
    u = !ii * 2 - 1;
    d0[ii] = inf * u;
    d1[ii] = -inf * u;
    d2[ii] = inf * u;
  }
  for (i = 0; i < n; i++) {
    cin >> s;
    ln = s.length();
    for (ii = 0; ii < 2; ii++) {
      fq[ii] = 0;
    }
    for (j = 0; j < ln; j++) {
      fq[s[j] == 'N']++;
    }
    for (ii = 0; ii < 2; ii++) {
      u = !ii * 2 - 1;
      d0[ii] = min(d0[ii] * u, fq[1] * u) * u;
      d1[ii] = max(d1[ii] * u, (fq[0] - fq[1]) * u) * u;
      d2[ii] = min(d2[ii] * u, fq[0] * u) * u;
    }
  }
  for (i = 0; i <= ma; i++) {
    lb = min(d0[0], i - d1[0]);
    rb = max(d0[1], i - d1[1]);
    ub = max(i - d2[0], d2[1] - i);
    p = max(lb + (rb - lb) / 2, (long long)!i);
    k = max(max(p - lb, rb - p), ub);
    if (k < z) {
      z = k;
      sq[0] = i;
      sq[1] = p;
    }
  }
  printf("%lld\n", z);
  for (ii = 0; ii < 2; ii++) {
    for (i = 0; i < sq[ii]; i++) {
      printf("%c", "BN"[ii]);
    }
  }
  printf("\n");
}