#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 998244353;
const int N = 1e6 + 10;
int power(int a, int b) {
  if (!b) return 1;
  int c = power(a, b / 2);
  c = (1LL * c * c) % mod;
  if (b % 2) c = (1LL * c * a) % mod;
  return c;
}
void solve() {
  int n, k;
  cin >> n >> k;
}
string s;
const string t = "RGB";
string ans;
int n;
int get(int x, int y) {
  int p = 0;
  string d = s;
  for (int i = 0; i < s.size(); ++i) {
    if (t[y % 3] != d[i]) ++p;
    d[i] = t[y % 3];
    ++y;
  }
  if (p < x) ans = d;
  return p;
}
int ch(int x, int y) {
  int p = 0;
  string d = s;
  for (int i = 0; i < n; ++i) {
    if (t[y % 3] != d[i]) ++p;
    d[i] = t[y % 3];
    --y;
    if (y < 0) y += 3;
  }
  if (p < x) ans = d;
  return p;
}
void another() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  bool ok = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (s[i + 1] < s[i]) {
      s.erase(s.begin() + i);
      ok = 1;
      break;
    }
  }
  if (!ok) {
    s.erase(s.begin() + (n - 1));
  }
  cout << s << '\n';
}
void test_case() {
  int t;
  cin >> t;
  while (t--) another();
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  another();
}

### Prompt

Generate a Cpp solution to the following problem:

You are given a string s consisting of n lowercase Latin letters.

You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.

String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p ≤ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.

For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of s.

The second line of the input contains exactly n lowercase Latin letters — the string s.

Output

Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.

Examples

Input


3
aaa


Output


aa


Input


5
abcda


Output


abca

Note

In the first example you can remove any character of s to obtain the string "aa".

In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 998244353;
const int N = 1e6 + 10;
int power(int a, int b) {
  if (!b) return 1;
  int c = power(a, b / 2);
  c = (1LL * c * c) % mod;
  if (b % 2) c = (1LL * c * a) % mod;
  return c;
}
void solve() {
  int n, k;
  cin >> n >> k;
}
string s;
const string t = "RGB";
string ans;
int n;
int get(int x, int y) {
  int p = 0;
  string d = s;
  for (int i = 0; i < s.size(); ++i) {
    if (t[y % 3] != d[i]) ++p;
    d[i] = t[y % 3];
    ++y;
  }
  if (p < x) ans = d;
  return p;
}
int ch(int x, int y) {
  int p = 0;
  string d = s;
  for (int i = 0; i < n; ++i) {
    if (t[y % 3] != d[i]) ++p;
    d[i] = t[y % 3];
    --y;
    if (y < 0) y += 3;
  }
  if (p < x) ans = d;
  return p;
}
void another() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  bool ok = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (s[i + 1] < s[i]) {
      s.erase(s.begin() + i);
      ok = 1;
      break;
    }
  }
  if (!ok) {
    s.erase(s.begin() + (n - 1));
  }
  cout << s << '\n';
}
void test_case() {
  int t;
  cin >> t;
  while (t--) another();
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  another();
}

```