/*
网络流牌大模拟！！！
黑白染色，黑色与白色是相反的，下面看黑色 (x+y)%2==1 的情况.
⚠ 四个位置对应 左，下，右，上.

单头型：
 	0001 (上,左)(上,右)1，(上,下)2.
 	0010 (右,上)(右,下)1，(右,左)2.
 	0100 (下,左)(下,右)1，(下,上)2.
	1000 (左,上)(左,下)1，(左,右)2.
直线型：
	1010 不做处理.
	0101 不做处理.
拐弯型：
	1100 (左,右)1，(下,上)1.
	0110 (右,左)1，(下,上)1.
	0011 (右,左)1，(上,下)1.
	1001 (左,右)1，(上,下)1.
三分型：
	1110 (左,上)(右,上)1，(下,上)2
	1101 (上,右)(下,右)1，(左,右)2
	1011 (左,下)(右,下)1，(上,下)2
	0111 (上,左)(下,左)1，(右,左)2
十字型：
	1111 不做处理.
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 114514, inf = 0x3f3f3f3f;
struct Edge {int now, nxt, w, cost;} e[1919810];
int head[N], cur = 1;
void ad(int u, int v, int w, int cost) {e[++cur].now = v, e[cur].nxt = head[u], e[cur].w = w, e[cur].cost = cost, head[u] = cur;}
void add(int u, int v, int w, int cost) {ad(u, v, w, cost), ad(v, u, 0, -cost);}
int s, t;
namespace G {
	int dis[N], icost[N], pre[N]; bool inque[N];
	bool spfa()
	{
		queue <int> q;
		memset(dis, 0x3f, sizeof dis), memset(icost, 0, sizeof icost);
		q.push(s), dis[s] = 0, icost[s] = inf, inque[s] = true;
		while (!q.empty())
		{
			int u = q.front(); q.pop(), inque[u] = false;
			for (int i = head[u]; i; i = e[i].nxt)
			{
				int v = e[i].now;
				if (e[i].w && dis[u] + e[i].cost < dis[v])
				{
					dis[v] = dis[u] + e[i].cost, pre[v] = i, icost[v] = min(icost[u], e[i].w);
					if (!inque[v]) q.push(v), inque[v] = true;
				}
			}
		}
		return icost[t] > 0;
	}
	pair <int, int> EK()
	{
		int cost = 0, ans = 0;
		while (spfa())
		{
			int w = icost[t]; cost += w, ans += w * dis[t];
			for (int i = t; i != s; i = e[pre[i] ^ 1].now) e[pre[i]].w -= w, e[pre[i] ^ 1].w += w;
		}
		return {cost, ans};
	}
}
const string dict[16] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m, flow;
int ME(int x, int y) {return ((x - 1) * m + y) + 0 * n * m;}
int UP(int x, int y) {return ((x - 1) * m + y) + 1 * n * m;}
int DW(int x, int y) {return ((x - 1) * m + y) + 2 * n * m;}
int LF(int x, int y) {return ((x - 1) * m + y) + 3 * n * m;}
int RH(int x, int y) {return ((x - 1) * m + y) + 4 * n * m;}
#define zlt(X, Y, P) add(X(x, y), Y(x, y), 1, P)
#define hxy(X, Y, P) add(Y(x, y), X(x, y), 1, P)
void solveBlack(int x, int y, string op)
{
	//黑点向四周建边
	add(s, ME(x, y), inf, 0); int dx, dy;
	dx = x - 1, dy = y; if (1 <= dx && dx <= n && 1 <= dy && dy <= m) add(UP(x, y), DW(dx, dy), 1, 0);
	dx = x + 1, dy = y; if (1 <= dx && dx <= n && 1 <= dy && dy <= m) add(DW(x, y), UP(dx, dy), 1, 0);
	dx = x, dy = y - 1; if (1 <= dx && dx <= n && 1 <= dy && dy <= m) add(LF(x, y), RH(dx, dy), 1, 0);
	dx = x, dy = y + 1; if (1 <= dx && dx <= n && 1 <= dy && dy <= m) add(RH(x, y), LF(dx, dy), 1, 0);

	//点向可以直接到达的周建边
	if (op[0] == '1') flow++, zlt(ME, LF, 0);
	if (op[1] == '1') flow++, zlt(ME, DW, 0);
	if (op[2] == '1') flow++, zlt(ME, RH, 0);
	if (op[3] == '1') flow++, zlt(ME, UP, 0);

	//单头型
	if (op == "0001") zlt(UP, LF, 1), zlt(UP, RH, 1), zlt(UP, DW, 2);
	if (op == "0010") zlt(RH, UP, 1), zlt(RH, DW, 1), zlt(RH, LF, 2);
	if (op == "0100") zlt(DW, LF, 1), zlt(DW, RH, 1), zlt(DW, UP, 2);
	if (op == "1000") zlt(LF, UP, 1), zlt(LF, DW, 1), zlt(LF, RH, 2);

	//直线型
	if (op == "1010") return;
	if (op == "0101") return;

	//拐弯型
	if (op == "1100") zlt(LF, RH, 1), zlt(DW, UP, 1);
	if (op == "0110") zlt(RH, LF, 1), zlt(DW, UP, 1);
	if (op == "0011") zlt(RH, LF, 1), zlt(UP, DW, 1);
	if (op == "1001") zlt(LF, RH, 1), zlt(UP, DW, 1);

	//三分型
	if (op == "1110") zlt(LF, UP, 1), zlt(RH, UP, 1), zlt(DW, UP, 2);
	if (op == "1101") zlt(UP, RH, 1), zlt(DW, RH, 1), zlt(LF, RH, 2);
	if (op == "1011") zlt(LF, DW, 1), zlt(RH, DW, 1), zlt(UP, DW, 2);
	if (op == "0111") zlt(UP, LF, 1), zlt(DW, LF, 1), zlt(RH, LF, 2);

	//十字型
	if (op == "1111") return;
}
void solveWhite(int x, int y, string op)
{
	add(ME(x, y), t, inf, 0);

	//点向可以直接到达的周建边
	if (op[0] == '1') flow++, hxy(ME, LF, 0);
	if (op[1] == '1') flow++, hxy(ME, DW, 0);
	if (op[2] == '1') flow++, hxy(ME, RH, 0);
	if (op[3] == '1') flow++, hxy(ME, UP, 0);

	//单头型
	if (op == "0001") hxy(UP, LF, 1), hxy(UP, RH, 1), hxy(UP, DW, 2);
	if (op == "0010") hxy(RH, UP, 1), hxy(RH, DW, 1), hxy(RH, LF, 2);
	if (op == "0100") hxy(DW, LF, 1), hxy(DW, RH, 1), hxy(DW, UP, 2);
	if (op == "1000") hxy(LF, UP, 1), hxy(LF, DW, 1), hxy(LF, RH, 2);

	//直线型
	if (op == "1010") return;
	if (op == "0101") return;

	//拐弯型
	if (op == "1100") hxy(LF, RH, 1), hxy(DW, UP, 1);
	if (op == "0110") hxy(RH, LF, 1), hxy(DW, UP, 1);
	if (op == "0011") hxy(RH, LF, 1), hxy(UP, DW, 1);
	if (op == "1001") hxy(LF, RH, 1), hxy(UP, DW, 1);

	//三分型
	if (op == "1110") hxy(LF, UP, 1), hxy(RH, UP, 1), hxy(DW, UP, 2);
	if (op == "1101") hxy(UP, RH, 1), hxy(DW, RH, 1), hxy(LF, RH, 2);
	if (op == "1011") hxy(LF, DW, 1), hxy(RH, DW, 1), hxy(UP, DW, 2);
	if (op == "0111") hxy(UP, LF, 1), hxy(DW, LF, 1), hxy(RH, LF, 2);

	//十字型
	if (op == "1111") return;
}
int main()
{
	scanf("%d%d", &n, &m), s = 0, t = N - 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1, x; j <= m; j++)
		{
			scanf("%d", &x);
			if ((i + j) & 1) solveBlack(i, j, dict[x]);
			else solveWhite(i, j, dict[x]);
		}
	pair <int, int> ans = G::EK();
	cout << (ans.first * 2 != flow ? -1 : ans.second);
	return 0;
}