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>
const int maxn = 1e5 + 10;
const int N = 4e5 + 10;
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qmul(long long p, long long q, long long mo) {
  long long f = 0;
  while (q) {
    if (q & 1) f = (f + p) % mo;
    p = (p + p) % mo;
    q >>= 1;
  }
  return f;
}
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p % 998244353;
    p = p * p % 998244353;
    q >>= 1;
  }
  return f;
}
int n, m, k, t, ok[26], du[26], vis[26];
vector<int> e[26];
char a[maxn];
bool flag;
void dfs(int rt) {
  if (flag) return;
  vis[rt] = -1;
  int i;
  for (i = 0; i <= (int)e[rt].size() - 1; i++) {
    int x = e[rt][i];
    if (vis[x] == -1) {
      flag = true;
      return;
    } else if (!vis[x])
      dfs(x);
  }
  vis[rt] = 1;
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= (int)n; i++) {
    scanf("%s", a);
    for (j = 0; a[j]; j++) {
      ok[a[j] - 'a'] = 1;
      if (a[j + 1]) e[a[j] - 'a'].push_back(a[j + 1] - 'a');
    }
  }
  for (i = 0; i <= (int)25; i++) {
    sort(e[i].begin(), e[i].end());
    e[i].erase(unique(e[i].begin(), e[i].end()), e[i].end());
    for (j = 0; j <= (int)e[i].size() - 1; j++) ++du[e[i][j]];
  }
  for (i = 0; i <= (int)25; i++)
    if (du[i] > 1 || e[i].size() > 1) return 0 * puts("NO");
  for (i = 0; i <= (int)25; i++)
    if (!vis[i]) dfs(i);
  if (flag) return 0 * puts("NO");
  for (i = 0; i <= (int)25; i++) {
    if (!du[i] && ok[i]) {
      j = i;
      putchar(j + 'a');
      while (e[j].size()) {
        j = e[j][0];
        putchar(j + 'a');
      }
    }
  }
  printf("\n");
  return 0;
}
