#include <iostream>
#include <vector>

using namespace std;

int numOfplan();

int main()
{
	cout << numOfplan() << endl;
	
	return 0;
}

int numOfplan()
{
	int n;
	cin >> n;

	vector<int> arr(n);
	int sum = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> arr[i];
		sum += arr[i];
	}

	if (sum % arr.size() != 0)
		return -1;

	int avg = sum / arr.size();
	int iCount = 0;
	for (unsigned int i = 0; i < arr.size(); i++)
	{
		if ((arr[i] - avg) % 2 == 0)
			iCount += (arr[i] - avg) > 0 ? (arr[i] - avg) / 2 : (avg - arr[i]) / 2;
		else
			return -1;
	}

	return iCount & 1 ? -1 : iCount / 2;
}

#include <vector>
#include <iostream>
#include <cmath>

using namespace std;

long long _maxWastage1(long long h);	// 方法1 
long long _maxWastage2(long long h);    // 方法2 

int main()
{
	// 由于 10^18 在 long long 的表示范围内

	long long h;
	cin >> h;

	cout << _maxWastage2(h) << endl;

	return 0;
}

long long _maxWastage1(long long h)
{
	long long result = sqrt(h);

	for (long long i = result - 1; i >= 0; i--)
	{
		if (i + i * i < h)
			return i;
	}

	return 0;
}

long long _maxWastage2(long long h)
{
	long long left = 0;
	long long right = h;
	long long mid = 0;

	while (left < right)
	{
		mid = (right + left + 1) >> 1;

		if (mid + mid * mid > h)
			right = mid - 1;
		else
			left = mid;
	}

	return left;
}

#include <iostream>

using namespace std;

bool is_ok_(int _m);
bool is_ok(int l, int r, int _m);

#define MAXN (500+10)  
#define MAXM (500+10)  
#define MAXT (2000000+10)  

int a[MAXN][MAXM], sum[MAXN][MAXM] = { 0 };
int n, m;

int main()
{
	cin >> n >> m;

	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			cin >> a[i][j];

			sum[i][j] = sum[i - 1][j] + a[i][j];
		}
	}

	int left = 1, right = 1, ans = 0;

	for (int j = 1; j <= m; j++)
		right += sum[n][j];

	for (int i = 1; i <= 60; i++)
	{
		int mid = (left + right) / 2;
		if (is_ok_(mid))
		{
			left = ans = mid;
		}
		else
			right = mid;
	}

	cout << ans << endl;

	return 0;
}

bool is_ok_(int _m)
{
	int p = 0, l = 1;

	for (int i = 1; i <= n; i++)
	{
		if (is_ok(l, i, _m))
		{
			l = i + 1; p++;
		}
	}

	if (p >= 4)
		return 1;
	else
		return 0;
}

bool is_ok(int l, int r, int _m)
{
	int tot = 0, p = 0;

	for (int i = 1; i <= m; i++)
	{
		tot += sum[r][i] - sum[l - 1][i];

		if (tot >= _m)
		{
			tot = 0;
			p++;
		}
	}

	if (p >= 4)
		return 1;
	else
		return 0;
}