#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll=long long;
using ld=long double;
ld pie=3.141592653589793;
ll inf=144499999999994;
ll mod=998244353;
int main(){
    ll n,m;
    cin >> n >> m;
    vector<vector<ll>>g(n+2,vector<ll>(n+2,inf));
    for (ll i = 1; i <=n; i++)
    {
        for (ll j = 1; j <=n; j++)
        {
            g[i][j]=0;
        }
    }
    for (ll i = 0; i < m; i++)
    {
        ll h,w,c;
        cin >> h >> w >> c;
        g[h][w]=c;
    }
    queue<pair<ll,ll>>que;
    que.push({1,1});
    vector<vector<ll>>memo1(n+2,vector<ll>(n+2,inf)),memo2(n+2,vector<ll>(n+2,inf)),memo(n+2,vector<ll>(n+2,0));
    memo1[1][1]=0;
    memo2[1][1]=0;
    while (!que.empty())
    {
        ll x=que.front().first;
        ll y=que.front().second;
        que.pop();
        if (x<=0||x>n||y<=0||y>n)
        {
            continue;
        }
        memo[x][y]=0;
        for (ll i = -1; i <=1 ; i++)
        {
            for (ll j = -1; j <=1; j++)
            {
                if (abs(i)+abs(j)!=1)
                {
                    continue;
                }
                if (memo1[x+i][y+j]>memo1[x][y]+g[x+i][y+j]+1)
                {
                    memo1[x+i][y+j]=memo1[x][y]+g[x+i][y+j]+1;
                    if (memo[x+i][y+j]==0)
                    {
                        que.push({x+i,y+j});
                        memo[x+i][y+j]=1;
                    }
                }
                if (memo2[x+i][y+j]>memo1[x][y]+1)
                {
                    memo2[x+i][y+j]=memo1[x][y]+1;
                    if (memo[x+i][y+j]==0)
                    {
                        que.push({x+i,y+j});
                        memo[x+i][y+j]=1;
                    }
                }
                if (memo2[x+i][y+j]>memo2[x][y]+g[x+i][y+j]+1)
                {
                    memo2[x+i][y+j]=memo2[x][y]+g[x+i][y+j]+1;
                    if (memo[x+i][y+j]==0)
                    {
                        que.push({x+i,y+j});
                        memo[x+i][y+j]=1;
                    }
                }
            }
        }
    }
    cout << min(memo1[n][n],memo2[n][n]) << endl;
}