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;
const int N = 3e5 + 11, M = 5e5 + 11, oo = ~0u >> 2;
char s[M];
int n, a[N], b[N];
int ansx, ansy;
int chk(int w) {
  int x = 0, y = 0;
  int X = oo, Y = oo;
  int u = -oo, v = oo;
  int i;
  for (i = 1; i <= n; ++i) {
    x = max(x, a[i] - w);
    y = max(y, b[i] - w);
    X = min(X, a[i] + w);
    Y = min(Y, b[i] + w);
    u = max(u, b[i] - a[i] - w);
    v = min(v, b[i] - a[i] + w);
  }
  if (x > X || y > Y || u > v) return 0;
  for (ansx = x; ansx <= X; ++ansx) {
    int yy = u + ansx;
    int YY = v + ansx;
    int _y = max(yy, y);
    int _Y = min(YY, Y);
    if (_y <= _Y && (_Y || ansx)) {
      ansy = _Y;
      return 1;
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  int i, j, mx = 0;
  for (i = 1; i <= n; ++i) {
    cin >> s;
    int l = strlen(s), t = 0;
    mx = max(mx, l);
    for (j = 0; j < l; ++j) {
      t += s[j] == 'B';
    }
    a[i] = t;
    b[i] = l - t;
  }
  int l = 0, r = mx + 1;
  while (l < r) {
    int m = l + r >> 1;
    if (chk(m))
      r = m;
    else
      l = m + 1;
  }
  printf("%d\n", l);
  chk(l);
  for (i = 1; i <= ansx; ++i) putchar('B');
  for (i = 1; i <= ansy; ++i) putchar('N');
  puts("");
}