//yukicoder@cpp17

#include <iostream>
#include <iomanip>

#include <algorithm>

#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>

#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>

#include <random>

#include <bitset>

#include <complex>

#include <utility>

#include <numeric>

#include <functional>


using namespace std;
using ll = long long;
using P = pair<ll,ll>;

const ll MOD = 998244353;
const ll MODx = 1000000007;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62LL)-1;
const double EPS = (1e-10);

P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};




template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }
 
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる

計算量は変わらないが楽できるシリーズ
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる
*/

/* comment outed because can cause bugs
__attribute__((constructor))
void initial() {
  cin.tie(0);
  ios::sync_with_stdio(false);
}
*/

template<typename T>
class dijkstra{
  struct vg{
    int t;
    T c;
  };
  vector<pair<int,T>> cost;
  int beg = -1;
  void run(int x){
    auto comp = [](vg a,vg b){
      return a.c>b.c;
    };
    priority_queue<vg,vector<vg>,decltype(comp)> R{comp};
    for(int i = 0; n > i; i++){
      cost[i].second = -1;
    }
    cost[x].first = -1;
    cost[x].second = 0;
    R.push({x,0});
    while(R.size()){
      auto k = R.top();R.pop();
      if(cost[k.t].second != k.c)continue;
      for(int i = 0; A[k.t].size() > i; i++){
        if(cost[A[k.t][i].t].second != -1 && cost[A[k.t][i].t].second <= k.c+A[k.t][i].c)continue;
        cost[A[k.t][i].t].second = k.c+A[k.t][i].c;
        cost[A[k.t][i].t].first = k.t;
        R.push({A[k.t][i].t,k.c+A[k.t][i].c});
      }
    }

  }

public:
  int n;
  vector<vector<vg>> A;
  vector<vector<vg>> Rev;
  dijkstra(int n_):n(n_),A(n_),Rev(n_),cost(n_,{-1,-1}){}

  //双方向
  void push(int s,int v,T c){
    A[s].push_back({v,c});
    A[v].push_back({s,c});
    Rev[s].push_back({v,c});
    Rev[v].push_back({s,c});
  }

  void push_side(int s,int v,T c){
    A[s].push_back({v,c});
    Rev[v].push_back({s,c});
  }

  void build(int x){
    run(x);
    beg = x;
  }

  //restorate any

  vector<int> restoration(int t = -1){
    vector<int> Ret;
    if(beg == -1 || (t != -1 && cost[t].second == -1)){
      Ret.push_back(-1);
      return Ret;
    }
    int ind = t;
    if(ind == -1){
      T tmp = cost[0].second;
      int tmpin = 0;
      for(int i = 0; n > i; i++){
        if(tmp < cost[i].second){
          tmp = cost[i].second;
          tmpin = i;
        }
      }
      ind = tmpin;
    }
    Ret.push_back(ind);
    while(beg != ind){
      Ret.push_back({cost[ind].first});
      ind = cost[ind].first;
    }
    reverse(Ret.begin(),Ret.end());
    return Ret;
  }

  T operator[](int i){
    return cost[i].second;
  }
};
int main(){
  int n,m;cin>>n>>m;
  vector<vector<int>> A(n, vector<int>(n));
  for(int i = 0; m > i; i++){
    int h,w,c;cin>>h>>w>>c;h--;w--;
    A[h][w] = c;
  }
  dijkstra<long long> D(n*n*2);
  for(int i = 0; n > i; i++){
    for(int j = 0; n > j; j++){
      if(i){
        D.push_side(       ((i-1)*n+j),  (      (i*n+j)), A[i][j]+1);
        D.push_side(       ((i-1)*n+j) , ((n*n)+(i*n+j)) , 1);
        D.push_side(((n*n)+((i-1)*n+j)), ((n*n)+(i*n+j)), A[i][j]+1);
      }
      if(i+1 != n){
        D.push_side(       ((i+1)*n+j),  (      (i*n+j)), A[i][j]+1);
        D.push_side(       ((i+1)*n+j) , ((n*n)+(i*n+j)) , 1);
        D.push_side(((n*n)+((i+1)*n+j)), ((n*n)+(i*n+j)), A[i][j]+1);
      }
      if(j){
        D.push_side(       (i*n+(j-1)),  (      (i*n+j)), A[i][j]+1);
        D.push_side(       (i*n+(j-1)) , ((n*n)+(i*n+j)) , 1);
        D.push_side(((n*n)+(i*n+(j-1))), ((n*n)+(i*n+j)), A[i][j]+1);
      }
      if(j+1 != n){
        D.push_side(       (i*n+(j+1)),  (      (i*n+j)), A[i][j]+1);
        D.push_side(       (i*n+(j+1)) , ((n*n)+(i*n+j)) , 1);
        D.push_side(((n*n)+(i*n+(j+1))), ((n*n)+(i*n+j)), A[i][j]+1);
      }
    }
  }
  D.build(0);
  cout << D[(n*n*2-1)] << endl;

  return 0;
}