#include <bits/stdc++.h>
using namespace std;
long long int M = 1000000007;
long long int i, j;
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
long long int mod(long long a, long long b) { return (a % b + b) % b; }
long long int N = 1000001;
vector<long long int> fcto(N);
void fac(long long int n) {
  fcto[1] = 1, fcto[0] = 1;
  for (long long int in = 2; in <= n; in++) {
    fcto[in] = in * fcto[in - 1] % M;
    fcto[in] %= M;
  }
}
long long int binexp(long long int a, long long int b) {
  if (a == 0 || a == 1) return a;
  if (b == 0) return 1;
  long long int ret = binexp(a, b / 2);
  if (b % 2)
    return (((ret * ret) % M) * a) % M;
  else
    return (ret * ret) % M;
}
long long int inv_fac(long long int x) { return binexp(fcto[x], M - 2); }
long long int ncr(long long int n, long long int r) {
  return ((fcto[n] * inv_fac(n - r) % M) * inv_fac(r)) % M;
}
long long int isPrime(long long int n) {
  if (n <= 3) return (1);
  if (n % 2 == 0 || n % 3 == 0) return (0);
  for (i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return (0);
  return (1);
}
long long int smallestDivisor(long long int n) {
  if (n % 2 == 0) return 2;
  for (i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return i;
  }
  return n;
}
long long int countDivisors(long long int n) {
  long long int cnt = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long long int maxSubArraySum(long long int a[], long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long int tt = 1;
  while (tt--) {
    long long int n;
    cin >> n;
    string s;
    cin >> s;
    long long int loc = -1;
    for (i = 0; i < n - 1; i++) {
      if (s[i] > s[i + 1]) {
        loc = i;
        break;
      }
    }
    if (loc == -1) {
      s.erase(n - 1, 1);
    } else {
      s.erase(loc, 1);
    }
    cout << s << "\n";
  }
}

### Prompt

Please formulate 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;
long long int M = 1000000007;
long long int i, j;
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
long long int mod(long long a, long long b) { return (a % b + b) % b; }
long long int N = 1000001;
vector<long long int> fcto(N);
void fac(long long int n) {
  fcto[1] = 1, fcto[0] = 1;
  for (long long int in = 2; in <= n; in++) {
    fcto[in] = in * fcto[in - 1] % M;
    fcto[in] %= M;
  }
}
long long int binexp(long long int a, long long int b) {
  if (a == 0 || a == 1) return a;
  if (b == 0) return 1;
  long long int ret = binexp(a, b / 2);
  if (b % 2)
    return (((ret * ret) % M) * a) % M;
  else
    return (ret * ret) % M;
}
long long int inv_fac(long long int x) { return binexp(fcto[x], M - 2); }
long long int ncr(long long int n, long long int r) {
  return ((fcto[n] * inv_fac(n - r) % M) * inv_fac(r)) % M;
}
long long int isPrime(long long int n) {
  if (n <= 3) return (1);
  if (n % 2 == 0 || n % 3 == 0) return (0);
  for (i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return (0);
  return (1);
}
long long int smallestDivisor(long long int n) {
  if (n % 2 == 0) return 2;
  for (i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return i;
  }
  return n;
}
long long int countDivisors(long long int n) {
  long long int cnt = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long long int maxSubArraySum(long long int a[], long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long int tt = 1;
  while (tt--) {
    long long int n;
    cin >> n;
    string s;
    cin >> s;
    long long int loc = -1;
    for (i = 0; i < n - 1; i++) {
      if (s[i] > s[i + 1]) {
        loc = i;
        break;
      }
    }
    if (loc == -1) {
      s.erase(n - 1, 1);
    } else {
      s.erase(loc, 1);
    }
    cout << s << "\n";
  }
}

```