#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define pll pair<ll,ll>
#define pii pair<int,int>
#define pq priority_queue
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define endl '\n'
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define lb(c,x) distance(c.begin(),lower_bound(all(c),x))
#define ub(c,x) distance(c.begin(),upper_bound(all(c),x))

using namespace std;

inline int topbit(unsigned long long x){
	return x?63-__builtin_clzll(x):-1;
}

inline int popcount(unsigned long long x){
	return __builtin_popcountll(x);
}

inline int parity(unsigned long long x){//popcount%2
	return __builtin_parity(x);
}



template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}

const ll INF=1e18;

const ll white=0; //未探索
const ll gray=1; //探索中
const ll black=2; //探索済み

vector<pll> g[500005]; 

ll v,r;

vector<ll> d;

void dijkstra(){
    priority_queue<pll> PQ;
    vector<ll> color(v);
    rep(i,v){
        d[i]=INF;
        color[i]=white;
    }
    d[r]=0;
    PQ.push(make_pair(0,r));
    color[r]=gray;
    while(!PQ.empty()){
        pll f=PQ.top();
        PQ.pop();
        ll u=f.second;
        color[u]=black;
        if(d[u]<f.first*(-1)) continue; //d[u]は正。
        for(ll j=0;j<g[u].size();j++){
            ll v=g[u][j].first;
            if(color[v]==black) continue;
            if(d[v]>d[u]+g[u][j].second){
                d[v]=d[u]+g[u][j].second;
                PQ.push(make_pair(d[v]*(-1),v));
                color[v]=gray;
            }
        }
    }
    
}

ll dx[4]={-1,1,0,0};
ll dy[4]={0,0,-1,1};

int main(){
    ll n,m;
    cin >> n >> m;
    vector<ll> h(m),w(m),c(m);
    map<pll,ll> e;
    rep(i,m){
        cin >> h[i] >> w[i] >> c[i];
        h[i]--;
        w[i]--;
        e[make_pair(h[i],w[i])]=c[i];
    }
    v=2*n*n;
    d=vector<ll>(v,INF);
    rep(i,n){
        rep(j,n){
            ll now=i*n+j;
            if(!e.count({i,j})){
                rep(k,4){
                    ll nx=i+dx[k];
                    ll ny=j+dy[k];
                    ll from=nx*n+ny;
                    if(0<=nx && nx<=n-1 && 0<=ny && ny<=n-1){
                        g[from].push_back({now,1LL});
                        g[from+n*n].push_back({now+n*n,1LL});
                    }
                }
            }
            else{
                ll cost=e[{i,j}];
                rep(k,4){
                    ll nx=i+dx[k];
                    ll ny=j+dy[k];
                    ll from=nx*n+ny;
                    if(0<=nx && nx<=n-1 && 0<=ny && ny<=n-1){
                        g[from].push_back({now,cost+1});
                        g[from+n*n].push_back({now+n*n,cost+1});
                        g[from].push_back({now+n*n,1});
                    }
                }
            }
        }
    }
    dijkstra();
    ll ans=INF;
    ans=min(d[n*n-1],d[2*n*n-1]);
    cout << ans << endl;
    return 0;
}