#include <bits/stdc++.h>
using namespace std;
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline int fastpo(int x, int n, int mod) {
  int res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
bool f[1111111];
int main() {
  for (int i(0); i <= 1000; i++) f[i * i] = true;
  int n;
  scanf("%d", &n);
  int ans = -inf;
  for (int i(1); i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x >= 0 && f[x]) {
    } else {
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
}

### Prompt

In CPP, your task is to solve the following problem:

Given an array a1, a2, ..., an of n integers, find the largest number in the array that is not a perfect square.

A number x is said to be a perfect square if there exists an integer y such that x = y2.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an ( - 106 ≤ ai ≤ 106) — the elements of the array.

It is guaranteed that at least one element of the array is not a perfect square.

Output

Print the largest number in the array which is not a perfect square. It is guaranteed that an answer always exists.

Examples

Input

2
4 2


Output

2


Input

8
1 2 4 8 16 32 64 576


Output

32

Note

In the first sample case, 4 is a perfect square, so the largest number in the array that is not a perfect square is 2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline int fastpo(int x, int n, int mod) {
  int res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
bool f[1111111];
int main() {
  for (int i(0); i <= 1000; i++) f[i * i] = true;
  int n;
  scanf("%d", &n);
  int ans = -inf;
  for (int i(1); i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x >= 0 && f[x]) {
    } else {
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
}

```