#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, long long power, long long mod) {
  if (!power) return 1;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
inline void run() {
  int arr[100];
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n;) scanf("%d", arr + i++);
  sort(arr, arr + n);
  if (arr[0] == arr[n - 1]) return (void)puts("NO");
  for (int i = 1;; ++i)
    if (arr[i] > arr[0]) return (void)printf("%d\n", arr[i]);
}
int main() {
  FILE* input = stdin;
  FILE* output = stdout;
  while (!feof(input)) {
    run();
    break;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.

Input

The first input line contains integer n (1 ≤ n ≤ 100) — amount of numbers in the sequence. The second line contains n space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.

Output

If the given sequence has the second order statistics, output this order statistics, otherwise output NO.

Examples

Input

4
1 2 2 -4


Output

1


Input

5
1 2 3 1 1


Output

2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, long long power, long long mod) {
  if (!power) return 1;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
inline void run() {
  int arr[100];
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n;) scanf("%d", arr + i++);
  sort(arr, arr + n);
  if (arr[0] == arr[n - 1]) return (void)puts("NO");
  for (int i = 1;; ++i)
    if (arr[i] > arr[0]) return (void)printf("%d\n", arr[i]);
}
int main() {
  FILE* input = stdin;
  FILE* output = stdout;
  while (!feof(input)) {
    run();
    break;
  }
  return 0;
}

```