Problem: You are given n words, each of which consists of lowercase alphabet letters. Each word contains at least one vowel. You are going to choose some of the given words and make as many beautiful lyrics as possible.

Each lyric consists of two lines. Each line consists of two words separated by whitespace. 

A lyric is beautiful if and only if it satisfies all conditions below. 

  * The number of vowels in the first word of the first line is the same as the number of vowels in the first word of the second line. 
  * The number of vowels in the second word of the first line is the same as the number of vowels in the second word of the second line. 
  * The last vowel of the first line is the same as the last vowel of the second line. Note that there may be consonants after the vowel. 



Also, letters "a", "e", "o", "i", and "u" are vowels. Note that "y" is never vowel.

For example of a beautiful lyric, 

"hello hellooowww" 

"whatsup yowowowow" 

is a beautiful lyric because there are two vowels each in "hello" and "whatsup", four vowels each in "hellooowww" and "yowowowow" (keep in mind that "y" is not a vowel), and the last vowel of each line is "o".

For example of a not beautiful lyric, 

"hey man"

"iam mcdic" 

is not a beautiful lyric because "hey" and "iam" don't have same number of vowels and the last vowels of two lines are different ("a" in the first and "i" in the second).

How many beautiful lyrics can you write from given words? Note that you cannot use a word more times than it is given to you. For example, if a word is given three times, you can use it at most three times.

Input

The first line contains single integer n (1 ≤ n ≤ 10^{5}) — the number of words.

The i-th of the next n lines contains string s_{i} consisting lowercase alphabet letters — the i-th word. It is guaranteed that the sum of the total word length is equal or less than 10^{6}. Each word contains at least one vowel.

Output

In the first line, print m — the number of maximum possible beautiful lyrics.

In next 2m lines, print m beautiful lyrics (two lines per lyric).

If there are multiple answers, print any.

Examples

Input


14
wow
this
is
the
first
mcdics
codeforces
round
hooray
i
am
proud
about
that


Output


3
about proud
hooray round
wow first
this is
i that
mcdics am


Input


7
arsijo
suggested
the
idea
for
this
problem


Output


0


Input


4
same
same
same
differ


Output


1
same differ
same same

Note

In the first example, those beautiful lyrics are one of the possible answers. Let's look at the first lyric on the sample output of the first example. "about proud hooray round" forms a beautiful lyric because "about" and "hooray" have same number of vowels, "proud" and "round" have same number of vowels, and both lines have same last vowel. On the other hand, you cannot form any beautiful lyric with the word "codeforces".

In the second example, you cannot form any beautiful lyric from given words.

In the third example, you can use the word "same" up to three times.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<V<T>>;
template <typename T>
using Mi = map<long long, T>;
template <typename T>
using MMi = map<long long, Mi<T>>;
const double PI = 3.141592653589793238463;
pair<long long, long long> zero = {-1, -1};
void cinArr(vector<long long> &arr) {
  for (int i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void cinArr(V<int> &arr) {
  for (int i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void cinMatr(vector<vector<long long>> &matr) {
  long long n = matr.size();
  long long m = matr[0].size();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> matr[i][j];
    }
  }
}
void coutArr(vector<long long> &arr) {
  for (long long val : arr) {
    cout << val << " ";
  }
  cout << '\n';
}
int xx[4] = {1, -1, 0, 0};
int yy[4] = {0, 0, 1, -1};
V<int> prim;
V<int> prims;
void er(long long n) {
  prim = V<int>(n + 1);
  for (int i = 0; i <= n; i++) {
    prim[i] = i;
  }
  for (long long i = 2; i <= n; i++) {
    if (prim[i] == i) {
      prims.push_back(i);
      for (long long j = i * i; j <= n; j += i) {
        if (prim[j] == j) {
          prim[j] = i;
        }
      }
    }
  }
}
class LCA {
 private:
  long long N;
  long long n;
  vector<vector<long long>> up;
  vector<vector<long long>> g;
  vector<long long> lvl;
  void dfs(long long v, long long h) {
    lvl[v] = h;
    for (long long u : g[v]) {
      dfs(u, h + 1);
    }
  }

 public:
  LCA(vector<long long> &parent) {
    n = parent.size();
    N = log2(n) + 1;
    g = vector<vector<long long>>(n);
    lvl = vector<long long>(n);
    long long core = -1;
    for (int i = 0; i < n; i++) {
      if (parent[i] != -1) {
        g[parent[i]].push_back(i);
      } else {
        core = i;
      }
    }
    dfs(core, 0);
    up = vector<vector<long long>>(N, vector<long long>(n));
    up[0] = parent;
    for (int i = 1; i < N; i++) {
      for (int j = 0; j < n; j++) {
        long long parent = up[i - 1][j];
        if (parent == -1) {
          up[i][j] = -1;
        } else {
          up[i][j] = up[i - 1][parent];
        }
      }
    }
  }
  long long getParent(long long v, long long needUp) {
    for (int i = N - 1; i >= 0; i--) {
      if (up[i][v] != -1 && lvl[v] - lvl[up[i][v]] <= needUp) {
        needUp -= lvl[v] - lvl[up[i][v]];
        v = up[i][v];
      }
    }
    if (needUp != 0) {
      return -1;
    } else {
      return v;
    }
  }
  long long getLCA(long long a, long long b) {
    if (lvl[a] > lvl[b]) {
      long long del = abs(lvl[a] - lvl[b]);
      a = getParent(a, del);
    }
    if (lvl[a] < lvl[b]) {
      long long del = abs(lvl[a] - lvl[b]);
      b = getParent(b, del);
    }
    if (a == b) {
      return a;
    }
    for (int i = N - 1; i >= 0; i--) {
      if (up[i][a] != up[i][b]) {
        a = up[i][a];
        b = up[i][b];
      }
    }
    return up[0][a];
  }
  long long getLvl(long long v) { return lvl[v]; }
};
class Node {
 private:
  long long a;
  long long del;
  LCA &lca;
  long long l;
  long long r;
  Node *lChildren;
  Node *rChildren;
  pair<long long, long long> merge(pair<long long, long long> x,
                                   pair<long long, long long> y, long long len1,
                                   long long len2) {
    if (len1 > 2 && len2 > 2) {
      long long a = x.first;
      long long b = x.second;
      long long c = y.first;
      long long d = y.second;
      long long ac = lca.getLCA(a, c);
      long long acb = lca.getLCA(ac, b);
      long long acd = lca.getLCA(ac, d);
      if (lca.getLvl(acb) > lca.getLvl(acd)) {
        return {acb, d};
      } else {
        return {acd, b};
      }
    }
    if (len1 == 0) {
      return y;
    }
    if (len2 == 0) {
      return x;
    }
    long long a = x.first;
    long long b = x.second;
    long long c = y.first;
    long long d = y.second;
    if (len1 == 1 && len2 == 1) {
      if (lca.getLvl(b) > lca.getLvl(d)) {
        return {b, d};
      } else {
        return {d, b};
      }
    }
    if (len1 == 1) {
      if (len2 == 2) {
        long long bc = lca.getLCA(b, c);
        long long bd = lca.getLCA(b, d);
        long long cd = lca.getLCA(c, d);
        long long lbc = lca.getLvl(bc);
        long long lbd = lca.getLvl(bd);
        long long lcd = lca.getLvl(cd);
        if (lbc > lbd && lbc > lcd) {
          return {bc, d};
        }
        if (lbd > lcd) {
          return {bd, c};
        }
        return {cd, b};
      }
      long long bc = lca.getLCA(b, c);
      long long cd = lca.getLCA(c, d);
      long long lbc = lca.getLvl(bc);
      long long lcd = lca.getLvl(cd);
      if (lbc > lcd) {
        return {bc, d};
      } else {
        return {cd, b};
      }
    }
    if (len2 == 1) {
      if (len1 == 2) {
        long long ab = lca.getLCA(a, b);
        long long ad = lca.getLCA(a, d);
        long long bd = lca.getLCA(b, d);
        long long lab = lca.getLvl(ab);
        long long lad = lca.getLvl(ad);
        long long lbd = lca.getLvl(bd);
        if (lab > lad && lab > lbd) {
          return {ab, d};
        }
        if (lad > lbd) {
          return {ad, b};
        }
        return {bd, a};
      }
      long long ab = lca.getLCA(b, a);
      long long ad = lca.getLCA(a, d);
      long long lab = lca.getLvl(ab);
      long long lad = lca.getLvl(ad);
      if (lab > lad) {
        return {ab, d};
      } else {
        return {ad, b};
      }
    }
    if (len1 == 2 && len2 > 2) {
      long long abc = lca.getLCA(a, lca.getLCA(b, c));
      long long acd = lca.getLCA(a, lca.getLCA(c, d));
      long long bcd = lca.getLCA(b, lca.getLCA(c, d));
      long long labc = lca.getLvl(abc);
      long long lacd = lca.getLvl(acd);
      long long lbcd = lca.getLvl(bcd);
      long long ans = max({labc, lacd, lbcd});
      if (labc == ans) {
        return {abc, d};
      }
      if (lacd == ans) {
        return {acd, b};
      }
      if (lbcd == ans) {
        return {bcd, a};
      }
    }
    if (len2 == 2 && len1 > 2) {
      long long abc = lca.getLCA(a, lca.getLCA(b, c));
      long long abd = lca.getLCA(a, lca.getLCA(b, d));
      long long acd = lca.getLCA(a, lca.getLCA(c, d));
      long long labc = lca.getLvl(abc);
      long long labd = lca.getLvl(abd);
      long long lacd = lca.getLvl(acd);
      long long ans = max({labc, labd, lacd});
      if (labc == ans) {
        return {abc, d};
      }
      if (labd == ans) {
        return {abd, c};
      }
      if (lacd == ans) {
        return {acd, b};
      }
    }
    long long abc = lca.getLCA(a, lca.getLCA(b, c));
    long long abd = lca.getLCA(a, lca.getLCA(b, d));
    long long acd = lca.getLCA(a, lca.getLCA(c, d));
    long long bcd = lca.getLCA(b, lca.getLCA(c, d));
    long long labc = lca.getLvl(abc);
    long long labd = lca.getLvl(abd);
    long long lacd = lca.getLvl(acd);
    long long lbcd = lca.getLvl(bcd);
    long long ans = max({labc, labd, lacd, lbcd});
    if (labc == ans) {
      return {abc, d};
    }
    if (labd == ans) {
      return {abd, c};
    }
    if (lacd == ans) {
      return {acd, b};
    }
    if (lbcd == ans) {
      return {bcd, a};
    }
    return zero;
  }
  Node(long long l, long long r, vector<long long> &arr, LCA &lca) : lca(lca) {
    this->l = l;
    this->r = r;
    if (l == r) {
      a = -1;
      del = arr[l];
    } else {
      long long mid = (l + r) / 2;
      lChildren = new Node(l, mid, arr, lca);
      rChildren = new Node(mid + 1, r, arr, lca);
      long long len1 = mid - l + 1;
      long long len2 = r - mid;
      pair<long long, long long> p =
          merge(lChildren->getValue(), rChildren->getValue(), len1, len2);
      a = p.first;
      del = p.second;
    }
  }

 public:
  Node(vector<long long> &arr, LCA &lca) : Node(0, arr.size() - 1, arr, lca) {}
  pair<long long, long long> getValue() {
    pair<long long, long long> p = {a, del};
    return p;
  }
  pair<long long, long long> getValue(long long left, long long right) {
    if (l == left && r == right) {
      return getValue();
    }
    long long mid = (l + r) / 2;
    pair<long long, long long> p1, p2;
    long long len1 = 0;
    long long len2 = 0;
    p1 = p2 = zero;
    if (left <= mid) {
      p1 = lChildren->getValue(left, min(right, mid));
      len1 = min(right, mid) - left + 1;
    }
    if (right > mid) {
      p2 = rChildren->getValue(max(mid + 1, left), right);
      len2 = right - max(mid + 1, left) + 1;
    }
    pair<long long, long long> p = merge(p1, p2, len1, len2);
    return p;
  }
  void update(long long l, long long r, long long value) {
    if (l == this->l && r == this->r) {
      return;
    }
    long long mid = (this->l + this->r) / 2;
    if (l <= mid) {
      lChildren->update(l, min(mid, r), value);
    }
    if (r > mid) {
      rChildren->update(max(mid + 1, l), r, value);
    }
  }
};
long long gcd(long long a, long long b) {
  if (a < b) {
    swap(a, b);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long nok(long long a, long long b) {
  long long nnod = gcd(a, b);
  long long ans = a / nnod * b;
  return ans;
}
long long divup(long long a, long long b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
class SNM {
 private:
  vector<long long> parent;
  vector<long long> size;

 public:
  SNM(long long n) {
    parent = vector<long long>(n);
    size = vector<long long>(n, 1);
    for (int i = 0; i < n; i++) {
      parent[i] = i;
    }
  }
  void make_set(int v) { parent[v] = v; }
  int find_set(int v) {
    if (v == parent[v]) return v;
    long long ans = find_set(parent[v]);
    parent[v] = ans;
    return ans;
  }
  void union_sets(int a, int b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
      parent[b] = a;
      size[a] += size[b];
    }
  }
  long long getSize(int v) {
    v = find_set(v);
    return size[v];
  }
};
long long bs(long long x, vector<long long> &arr) {
  long long l = 0;
  long long r = arr.size();
  while (l < r) {
    long long mid = (l + r) / 2;
    if (arr[mid] > x) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  return l;
}
void delR(long long v, vector<vector<long long>> &g, vector<bool> &viz,
          vector<vector<long long>> &G) {
  viz[v] = true;
  for (long long u : g[v]) {
    if (!viz[u]) {
      G[v].push_back(u);
      delR(u, g, viz, G);
    }
  }
}
pair<long long, long long> drob(long long a, long long b) {
  long long nod = gcd(a, b);
  a /= nod;
  b /= nod;
  return {a, b};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<bool> flags(26, false);
  flags[0] = true;
  flags['e' - 'a'] = true;
  flags['o' - 'a'] = true;
  flags['i' - 'a'] = true;
  flags['u' - 'a'] = true;
  V<string> arr(n);
  set<pair<long long, long long>> buffer;
  map<long long, map<long long, V<long long>>> strs;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    long long lastG;
    long long len = 0;
    for (int j = 0; j < arr[i].size(); j++) {
      if (flags[arr[i][j] - 'a']) {
        lastG = arr[i][j] - 'a';
        len++;
      }
    }
    strs[len][lastG].push_back(i);
  }
  set<pair<long long, long long>> alsoBuf;
  for (auto &C : strs) {
    auto &v = C.second;
    vector<long long> lenbuf;
    for (auto &Ms : v) {
      auto &ms = Ms.second;
      for (int i = 0; i < ms.size() - 1; i += 2) {
        long long s1 = ms[i];
        long long s2 = ms[i + 1];
        buffer.insert({s1, s2});
      }
      if (ms.size() % 2 != 0) {
        lenbuf.push_back(ms[ms.size() - 1]);
      }
    }
    long long len = lenbuf.size() / 2;
    for (int i = 0; i / 2 < len; i += 2) {
      long long s1 = lenbuf[i];
      long long s2 = lenbuf[i + 1];
      alsoBuf.insert({s1, s2});
    }
  }
  V<pair<pair<long long, long long>, pair<long long, long long>>> ans;
  while (buffer.size() > 0) {
    pair<long long, long long> p1 = *buffer.begin();
    buffer.erase(buffer.begin());
    if (alsoBuf.size() > 0) {
      pair<long long, long long> p2 = *alsoBuf.begin();
      alsoBuf.erase(alsoBuf.begin());
      ans.push_back({p2, p1});
    } else if (buffer.size() > 0) {
      pair<long long, long long> p2 = *buffer.begin();
      buffer.erase(buffer.begin());
      ans.push_back({p2, p1});
    } else {
      break;
    }
  }
  cout << ans.size() << '\n';
  for (auto pp : ans) {
    long long a = pp.first.first;
    long long b = pp.second.first;
    long long c = pp.first.second;
    long long d = pp.second.second;
    cout << arr[a] << " " << arr[b] << '\n' << arr[c] << " " << arr[d] << '\n';
  }
  return 0;
}