#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int inf = 0x3f3f3f3f; ///1061109567
const int maxn = 1e6 + 10;

int n, m;
int all;
int f[maxn][10];
vector<string> s;

///计算当前当前列到state状态需要的花费
int cal(int p, int state) {
    bitset<3> bt(state);
    int cnt = 0;
    for (int i = 0; i < n; ++ i) {
        if (bt[i] != s[i][p]-‘0‘) cnt ++;
    }
    return cnt;
}

///判断是否可以由st1转移到st2(判断合起来是否都是奇数1
bool check(int st1, int st2) {
    for (int i = 1; i < n; ++ i) {
        int cnt = ((st1>>i)&1) +((st1>>(i-1))&1) + ((st2>>i)&1) +((st2>>(i-1))&1);
        if (cnt % 2 == 0) return 0;
    }
    return 1;
}

///在前面所有可以转移的状态里面取小
int premin(int j, int state) {
    int minn = 1e9;
    for (int pre = 0; pre < all; ++ pre) {
        if (check(pre, state))
            minn = min(minn, f[j-1][pre]);
    }
    return minn;
}

int getans(int n) {
    if (n >= 4) return -1;  ///性质
    if (n == 1) return 0; 
    all = 1 << n;
    ///初始化
    for (int i = 0; i < all; ++ i) {
        f[0][i] = cal(0, i);
    }
    ///dp
    for (int j = 1; j < m; ++ j) {
        for (int i = 0; i < all; ++ i) {
            f[j][i] = premin(j, i) + cal(j, i);
        }
    }
    int ans = 1e9;
    for (int i = 0; i < all; ++ i)
        ans = min(ans, f[m-1][i]);
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++ i) {
        string t; cin >> t;
        s.push_back(t);
    }
    cout << getans(n) << endl;
    return 0;
}