#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <algorithm>
#include <bitset>
#include <climits>
#include <cmath>
#include <cstring>
#include <deque>
#include <forward_list>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;
typedef long long ll;
typedef pair<ll, pair<int, int>> P;
typedef bitset<16> BS;
struct edge {
    int to, cost;
};
const ll MOD = 1E+09 + 7;
const ll INF = 1E18;
const int MAX_N = 500;

int N, M;

ll d1[MAX_N + 1][MAX_N + 1], d2[MAX_N + 1][MAX_N + 1];
ll C[MAX_N + 1][MAX_N + 1];
ll dx[4] = { -1, 1, 0, 0 }, dy[4] = { 0, 0, -1, 1 };
int H[MAX_N * MAX_N], W[MAX_N * MAX_N];

void dijkstra(int si, int sj, ll d[][MAX_N + 1]);

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N >> M;

    for (int i = 1; i <= M; i++) {
        ll c;
        cin >> H[i] >> W[i] >> c;
        C[H[i]][W[i]] = c;
    }

    fill(d1[1], d1[N + 1], INF);
    fill(d2[1], d2[N + 1], INF);
    dijkstra(1, 1, d1);
    dijkstra(N, N, d2);
    ll ans = d1[N][N];
    for (int i = 1; i <= M; i++) {
        //cout << "H = " << H[i] << ", W = " << W[i] << ", c = " << d1[H[i]][W[i]] + d2[H[i]][W[i]] - 2 * C[H[i]][W[i]] << "\n";
        ans = min(ans, d1[H[i]][W[i]] + d2[H[i]][W[i]] - 2 * C[H[i]][W[i]]);
    }

    cout << ans << "\n";

    return 0;
}

void dijkstra(int si, int sj, ll d[][MAX_N + 1])
{
    priority_queue<P, vector<P>, greater<P>> que;
    que.push(P(0, make_pair(si, sj)));
    d[si][sj] = 0;
    while (!que.empty()) {
        P p = que.top();
        int vi = p.second.first;
        int vj = p.second.second;
        que.pop();
        if (d[vi][vj] < p.first)
            continue; //被った頂点に関しては更新前の古いデータはとばす
        for (int i = 0; i < 4; i++) {
            int ti = vi + dx[i], tj = vj + dy[i];
            if (ti >= 1 && ti <= N && tj >= 1 && tj <= N && d[ti][tj] > d[vi][vj] + 1 + C[ti][tj]) {
                d[ti][tj] = d[vi][vj] + 1 + C[ti][tj];
                que.push(P(d[ti][tj], make_pair(ti, tj)));
            }
        }
    }
}