Problem: You are given a string s of length n consisting of lowercase Latin letters. You may apply some operations to this string: in one operation you can delete some contiguous substring of this string, if all letters in the substring you delete are equal. For example, after deleting substring bbbb from string abbbbaccdd we get the string aaccdd.

Calculate the minimum number of operations to delete the whole string s.

Input

The first line contains one integer n (1 ≤ n ≤ 500) — the length of string s.

The second line contains the string s (|s| = n) consisting of lowercase Latin letters.

Output

Output a single integer — the minimal number of operation to delete string s.

Examples

Input


5
abaca


Output


3

Input


8
abcddcba


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int setbit(int n, int pos) { return n = n | (1 << pos); }
int resetbit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkbit(int n, int pos) { return (bool)(n & (1 << pos)); }
template <typename T>
T mul(T x, T y, T mod = 1e9 + 7) {
  x %= mod;
  y %= mod;
  return (x * y) % mod;
}
template <typename T>
T add(T x, T y, T mod = 1e9 + 7) {
  x %= mod;
  y %= mod;
  return (x + y) % mod;
}
template <typename T>
T sub(T x, T y, T mod = 1e9 + 7) {
  x %= mod;
  y %= mod;
  return ((x - y) % mod + mod) % mod;
}
template <typename T>
T ext_gcd(T num1, T num2, T &X, T &Y) {
  T r2, r1, q2, q1, x2, x1, y2, y1, x, y, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = num1, r1 = num2; r1 != 0;
       y2 = y1, y1 = y, x2 = x1, x1 = x, r2 = r1, r1 = r) {
    q1 = r2 / r1;
    x = x2 - q1 * x1;
    y = y2 - q1 * y1;
    r = r2 % r1;
  }
  X = x2;
  Y = y2;
  return r2;
}
template <typename T>
void print(const T &v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1 &first, const T2 &...rest) {
  print(first);
  print(rest...);
}
template <class T>
string to_str(T t) {
  stringstream second;
  second << t;
  return second.str();
}
template <typename T>
T bigmod(T x, T k, T mod = 1e9 + 7) {
  if (k == 0) return (T)1;
  if (k & 1) {
    return ((x % mod) * bigmod(x, k - 1, mod) % mod) % mod;
  } else {
    T ret = bigmod(x, k / 2, mod);
    ret %= mod;
    return (ret * ret) % mod;
  }
}
template <typename T>
T invmod(T x, T m) {
  return bigmod(x, m - 2, m);
}
template <typename T>
void dbg_a(T a[], int n = 10) {
  cerr << "[ ";
  for (int i = 0; i < n; i++) cerr << a[i] << ' ';
  cerr << " ]"
       << "\n";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "[ " << p.first << ", " << p.second << " ]";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
  return os << " ]\n";
}
template <typename T>
ostream &operator<<(ostream &os, const map<T, T> &Map) {
  os << "[ ";
  for (__typeof((Map).begin()) it = (Map).begin(); it != (Map).end(); ++it)
    os << "[" << (*it).first << ' ' << (*it).second << "] ";
  return os << "]\n";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &Set) {
  os << "[ ";
  for (__typeof((Set).begin()) it = (Set).begin(); it != (Set).end(); ++it)
    os << *it << ' ';
  return os << " ]\n";
}
const long long maxn = (long long)1e6 + 123;
const long long inf = 1e9 + 123;
const long long mod = 1e9 + 7;
long long dp[502][502];
string s;
long long func(long long l, long long r) {
  if (l > r) return 0;
  if (l == r) return 1;
  long long &ans = dp[l][r];
  if (ans != -1) return ans;
  ans = 1 + func(l + 1, r);
  for (long long k = l + 1; k <= r; k++) {
    if (s[k] == s[l]) {
      ans = min(ans, func(l + 1, k - 1) + func(k, r));
    }
  }
  return ans;
}
void solve() {
  long long l, r, n, k, tot = 0, x, y, ans = 0;
  cin >> n >> s;
  memset(dp, -1, sizeof dp);
  cout << func(0, n - 1) << "\n";
}
signed main() {
  clock_t st = clock();
  long long c = 0, t = 1;
  while (t--) {
    solve();
  }
  return 0;
}