Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.

The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.

Output

Print a single integer — the minimum number of seconds needed to destroy the entire line.

Examples

Input

3
1 2 1


Output

1


Input

3
1 2 3


Output

3


Input

7
1 4 4 2 3 2 1


Output

2

Note

In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long int> prime;
void sieve(long long int n) {
  bool bakh[n + 1];
  memset(bakh, true, sizeof(bakh));
  for (long long int p = 2; p * p <= n; p++) {
    if (bakh[p] == true) {
      for (long long int i = p * p; i <= n; i += p) bakh[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (bakh[p]) prime.push_back(p);
}
long long int eulertotient(long long int z) {
  long long int fac = z;
  for (long long int i = 0; prime[i] * prime[i] <= z; i++) {
    if (z % prime[i] == 0) {
      fac -= (fac / prime[i]);
      while (z % prime[i] == 0) z /= prime[i];
    }
  }
  if (z > 1) fac -= (fac / z);
  return fac;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  long long int ans = (a * b) / g;
  return ans;
}
long long int modInverse(long long int a, long long int m) {
  long long int hvf = gcd(a, m);
  if (hvf == 1) return power(a, m - 2, m);
  return -1;
}
void multiply(long long int F[2][2], long long int M[2][2]) {
  long long int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
  long long int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
  long long int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
  long long int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
void powermat(long long int F[2][2], long long int n) {
  if (n == 0 || n == 1) return;
  long long int M[2][2] = {{1, 1}, {1, 0}};
  powermat(F, n / 2);
  multiply(F, F);
  if (n % 2 != 0) multiply(F, M);
}
long long int fib(long long int n) {
  long long int F[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  powermat(F, n - 1);
  return F[0][0];
}
vector<long long int> merge(vector<long long int> v1,
                            vector<long long int> v2) {
  vector<long long int> temp;
  long long int i = 0, j = 0;
  while (i < v1.size() && j < v2.size()) {
    if (v1[i] < v2[j])
      temp.push_back(v1[i]), i++;
    else
      temp.push_back(v2[j]), j++;
  }
  while (i < v1.size()) temp.push_back(v1[i]), i++;
  while (j < v2.size()) temp.push_back(v2[j]), j++;
  return temp;
}
long long int n, dp[502][502];
vector<long long int> ar;
long long int fun(long long int i, long long int j) {
  if (i > j) return 0;
  if (i == j) return 1;
  if (i + 1 == j) return (ar[i] == ar[j] ? 1 : 2);
  if (dp[i][j] != -1) return dp[i][j];
  long long int p = INT_MAX;
  p = min(p, 1 + fun(i + 1, j));
  if (ar[i] == ar[j]) p = min(p, fun(i + 1, j - 1));
  if (ar[i] == ar[i + 1]) p = min(p, 1 + fun(i + 2, j));
  for (long long int k = i + 2; k < j; k++) {
    if (ar[i] == ar[k]) {
      p = min(p, fun(i + 1, k - 1) + fun(k + 1, j));
    }
  }
  return dp[i][j] = p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  ar.resize(n + 1, 0);
  for (long long int i = 1; i <= n; i++) cin >> ar[i];
  memset(dp, -1, sizeof(dp));
  long long int ans = fun(1, n);
  cout << ans << endl;
}
