#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define pb push_back
#define all(x) x.begin(),x.end()
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>
#define vvll vector<vll>
#define vs vector<string>
#define vvs vector<vs>
#define vc vector<char>
#define vvc vector<vc>
#define pii pair<int,int>
#define pllll pair<ll,ll>
#define vpii vector<pair<int,int>>
#define vpllll vector<pair<ll,ll>>
#define vpis vector<pair<int,string>>
#define vplls vector<pair<ll, string>>
#define vpsi vector<pair<string, int>>
#define vpsll vector<pair<string, ll>>

template<typename T>
bool chmax(T &a, const T &b) {bool flag = false; if(a < b) flag = true; a = (a > b? a : b); return flag;}
template<typename T>
bool chmin(T &a, const T &b) {bool flag = false; if(a > b) flag = true; a = (a < b? a : b); return flag;}

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 998244353;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};

#define int long long

#define vvvi vector<vvi>

void solve() {
    int n, m;
    cin >> n >> m;
    vvi a(n, vi(m));
    rep(i, n) rep(j, m) cin >> a[i][j];

    vvvi dp(n, vvi(m, vi(2, INF)));
    FOR(m) dp[0][i][0] = 0;
    //dp[i][j][k] -- (i,j)まで行くときそこが取り壊されてるかどうかがkであるときの最小コスト
    repi(i, 1, n) {
        int mn = INF;
        rep(j, m) {
            chmin(dp[i][j][1], dp[i-1][j][1] + a[i][j]);
            chmin(dp[i][j][1], dp[i-1][j][0] + a[i][j] + a[i-1][j]);
        }

        vector<int> order(m);
        iota(all(order), 0);
        sort(all(order), [&](int zi, int zj) {
            return dp[i][zi][1] < dp[i][zj][1];
        });

        rep(j, m) {
            if(j != order[0]) {
                chmin(dp[i][j][0], dp[i][order[0]][1]);
            }else {
                chmin(dp[i][j][0], dp[i][order[1]][1]);
            }
        }
    }

    int ans = INF;
    FOR(m) {
        chmin(ans, dp[n-1][i][0]);
        chmin(ans, dp[n-1][i][1]);
    }
    cout << ans << endl;
}

signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}