A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.

You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).

A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.

The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.

String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.

Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.

The total length of the strings doesn't exceed 105.

Output

Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.

Examples

Input

4
mail
ai
lru
cf


Output

cfmailru


Input

3
kek
preceq
cheburek


Output

NO

Note

One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
#include <bits/stdc++.h>
using namespace std;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) {
  long long z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
int pre[33], nxt[33];
char s[100005];
bool vis[33];
void fmain(int ID) {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (int)(30); (i)++) {
    pre[i] = -1;
    nxt[i] = -1;
  }
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%s", s);
    int len = strlen(s);
    for (int(j) = 0; (j) < (int)(len); (j)++) {
      int num = s[j] - 'a';
      vis[num] = 1;
      if (j) {
        int p = s[j - 1] - 'a';
        if (nxt[p] != -1 && nxt[p] != num) etp();
        nxt[p] = num;
        if (pre[num] != -1 && pre[num] != p) etp();
        pre[num] = p;
      }
    }
  }
  vector<string> vs;
  for (int(j) = 0; (j) < (int)(26); (j)++)
    if (vis[j] && pre[j] == -1) {
      string tmp = "";
      tmp += ('a' + j);
      vis[j] = 0;
      int t = j;
      while (nxt[t] != -1) {
        t = nxt[t];
        tmp += ('a' + t);
        vis[t] = 0;
      }
      vs.push_back(tmp);
    }
  for (int(j) = 0; (j) < (int)(26); (j)++)
    if (vis[j]) etp();
  sort(vs.begin(), vs.end());
  for (string o : vs) cout << o;
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
