While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.

Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.

Input

The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got. 

Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.

Output

If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO". 

If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.

Examples

Input

5
aca
aba
aba
cab
bac


Output

YES
abacaba


Input

4
abc
bCb
cb1
b13


Output

NO


Input

7
aaa
aaa
aaa
aaa
aaa
aaa
aaa


Output

YES
aaaaaaaaa
import java.util.*;
import java.io.*;
public class D {
  static int n;
  static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

  public static void main(String[] args) throws IOException {
    n = Integer.parseInt(br.readLine());

    if (n == 1) {
      System.out.println("YES");
      System.out.println(br.readLine());
      return;
    }

    int id = 0;
    Map<String, Integer> map = new HashMap<>();
    Map<Integer, String> invmap = new HashMap<>();
    String[] strings = new String[n];

    for (int i = 0; i < n; i++) {
      String s = br.readLine();
      strings[i] = s;

      String prefix = s.substring(0, 2);
      String suffix = s.substring(1, 3);

      if (!map.containsKey(prefix)) {
        map.put(prefix, id);
        invmap.put(id, prefix);
        id++;
      }

      if (!map.containsKey(suffix)) {
        map.put(suffix, id);
        invmap.put(id, suffix);
        id++;
      }
    }

    // Construct graph.
    int sz = map.size();
    List<List<Integer>> g = initializeEmptyGraph(sz);
    for (int i = 0; i < n; i++) {
      String s = strings[i];
      String prefix = s.substring(0, 2);
      String suffix = s.substring(1, 3);      
      int node1 = map.get(prefix);
      int node2 = map.get(suffix);
      addDirectedEdge(g, node1, node2);
    }

    EulerianPathDirectedEdgesAdjacencyList solver;
    solver = new EulerianPathDirectedEdgesAdjacencyList(g);

    int[] path = solver.getEulerianPath();
    if (path == null) {
      System.out.println("NO");
    } else {
      System.out.println("YES");
      StringBuilder sb = new StringBuilder();
      sb.append(invmap.get(path[0]));
      for (int i = 1; i < path.length; i++)
        sb.append(invmap.get(path[i]).charAt(1));
      System.out.println(sb);
    }

  }

  public static List<List<Integer>> initializeEmptyGraph(int n) {
    List<List<Integer>> graph = new ArrayList<>(n);
    for (int i = 0; i < n; i++) 
      graph.add(new ArrayList<>());
    return graph;
  }

  public static void addDirectedEdge(List<List<Integer>> g, int from, int to) {
    g.get(from).add(to);
  }

}

class EulerianPathDirectedEdgesAdjacencyList {
  
  public int n, edgeCount, orderIndex;
  public int[] in, out, ordering;
  public List<List<Integer>> graph;

  public EulerianPathDirectedEdgesAdjacencyList(List<List<Integer>> graph) {
    if (graph == null) throw new IllegalArgumentException("Graph cannot be null");
    n = graph.size();
    this.graph = graph;
  }

  private void setUp() {
    // Arrays that track the in degree and out degree of each node.
    in  = new int[n];
    out = new int[n];

    edgeCount = 0;

    // Compute in and out node degrees.
    for (int from = 0; from < n; from++) {
      for (int to : graph.get(from)) {
        in[to]++;
        out[from]++;
        edgeCount++;
      }
    }
  }

  private boolean graphHasEulerianPath() {
    int endNodes = 0, startNodes = 0;
    for (int i = 0; i < n; i++) {
      if (out[i] - in[i] > 1) // equivalently, in[i] - out[i] > 1 by symmetry.
        return false;
      if (in[i] - out[i] == 1) {
        endNodes++;
      } else if (out[i] - in[i] == 1) {
        startNodes++;
      }
    }
    return (endNodes == 0 && startNodes == 0) || 
           (endNodes == 1 && startNodes == 1);
  }

  private int findStartNode() {
    int start = 0;
    for (int i = 0; i < n; i++) {
      // Unique starting node.
      if (out[i] - in[i] == 1) return i;

      // Start at a node with an outgoing edge.
      if (out[i] > 0) start = i;
    }
    return start;
  }

  // Returns a list of edgeCount + 1 node ids that give the Eulerian path or
  // null if no path exists or the graph is disconnected.
  public int[] getEulerianPath() {
    setUp();

    if (edgeCount == 0) return null;
    if (!graphHasEulerianPath()) return null;

    orderIndex = edgeCount;
    ordering = new int[edgeCount+1];
    dfs(findStartNode());

    if (!graphIsConnected()) return null;

    return ordering;
  }

  // Perform DFS to find Eulerian path.
  private void dfs(int at) {
    while(out[at] != 0) {
      int next = graph.get(at).get(--out[at]);
      dfs(next);
    }
    ordering[orderIndex--] = at;
  }

  // Make sure all edges of the graph were traversed. It could be the case that
  // the graph is disconnected in which case return false.
  private boolean graphIsConnected() {
    for (int i = 0; i < n; i++)
      if (out[i] != 0)
        return false;
    return true;
  }

}









