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>
const int mod = 200003;
const int gmod = 2;
const int inf = 1039074182;
const long long llinf = 1LL * inf * inf;
template <typename T1, typename T2>
inline void chmin(T1 &x, T2 b) {
  if (b < x) x = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &x, T2 b) {
  if (b > x) x = b;
}
template <typename T1, typename T2>
inline void chadd(T1 &x, T2 b) {
  x += b;
  if (x >= mod) x -= mod;
}
template <typename T1, typename T2>
inline void chmul(T1 &x, T2 b) {
  x = 1LL * x * b % mod;
}
template <typename T1, typename T2>
inline void chmod(T1 &x, T2 b) {
  x %= b, x += b;
  if (x >= b) x -= b;
}
using namespace std;
template <typename T>
ostream &operator<<(ostream &cout, vector<T> vec) {
  cout << "{";
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i];
    if (i != (int)vec.size() - 1) cout << ',';
  }
  cout << "}";
  return cout;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &cout, pair<T1, T2> p) {
  cout << "(" << p.first << ',' << p.second << ")";
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, set<T> s) {
  vector<T> t;
  for (auto x : s) t.push_back(x);
  cout << t;
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, queue<T> q) {
  vector<T> t;
  while (q.size()) {
    t.push_back(q.front());
    q.pop();
  }
  cout << t;
  return cout;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &cout, map<T1, T2> m) {
  for (auto &x : m) {
    cout << "Key: " << x.first << ' ' << "Value: " << x.second << endl;
  }
  return cout;
}
template <typename T>
T operator*(vector<T> v1, vector<T> v2) {
  assert(v1.size() == v2.size());
  int n = v1.size();
  T res = 0;
  for (int i = 0; i < n; i++) {
    res += v1[i] * v2[i];
  }
  return res;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> x, pair<T1, T2> y) {
  return make_pair(x.first + y.first, x.second + y.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> x) {
  return make_pair(-x.first, -x.second);
}
template <typename T>
vector<vector<T> > operator~(vector<vector<T> > vec) {
  vector<vector<T> > v;
  int n = vec.size(), m = vec[0].size();
  v.resize(m);
  for (int i = 0; i < m; i++) {
    v[i].resize(n);
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      v[i][j] = vec[j][i];
    }
  }
  return v;
}
void print0x(int x) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  std::reverse(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
template <typename T>
void print0x(T x, int len) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  reverse(vec.begin(), vec.end());
  for (int i = vec.size(); i < len; i++) {
    putchar('0');
  }
  for (int i = 0; i < vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
using namespace std;
const int basic = 500000;
int n;
int a[300005], b[300005];
char buf[500005];
vector<int> sum[basic * 2 + 5];
int m1, m2, m3, m4;
int b1, b2;
int besta, bestb;
int best;
inline void calc(int k) {
  int res = 0;
  chmax(res, b1);
  chmax(res, b2);
  int zx = max(m2, m3);
  int nx = max(m1, m4);
  int tx = (nx - zx) / 2;
  chmax(tx, 0);
  chmax(res, -tx + nx);
  if (res < best) {
    best = res;
    besta = tx;
    if (besta == 0 && k == 0) besta++;
    bestb = besta + k;
  }
}
inline void read(int &a, int &b) {
  a = 0;
  b = 0;
  char c = 0;
  while (c != 'N' && c != 'B') c = getchar();
  while (c == 'N' || c == 'B') {
    if (c == 'B')
      b++;
    else
      a++;
    c = getchar();
  }
}
int main() {
  cin >> n;
  m1 = -inf, m2 = -inf, m3 = -inf, m4 = -inf;
  b1 = -inf, b2 = -inf;
  int k = basic;
  for (int i = 0; i < n; i++) {
    read(a[i], b[i]);
    sum[b[i] - a[i] + basic].push_back(i);
    chmax(b2, a[i] + k - b[i]);
    chmax(m3, k - b[i]);
    chmax(m4, a[i]);
  }
  best = inf;
  calc(basic);
  for (k = basic - 1; k >= -basic; k--) {
    m1++;
    b1++;
    m3--;
    b2--;
    for (auto id : sum[k + basic]) {
      int x0 = a[id];
      int y0 = b[id];
      chmax(m1, y0 - k);
      chmax(m2, -x0);
      chmax(b1, y0 - (x0 + k));
    }
    calc(k);
  }
  cout << best << endl;
  for (int i = 0; i < bestb; i++) {
    putchar('B');
  }
  for (int i = 0; i < besta; i++) {
    putchar('N');
  }
  return 0;
}