// Original C++ solution (slower):
#include <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<(n);++i)

using namespace std;

typedef long long ll;



const ll INF = 1LL << 60;



vector< vector< pair<int,int> > > to(100005);

ll dist[100005];



int main() {

  int N,M;

  cin >> N >> M;

  REP(i,M) {

    int l,r,c;

    cin >> l >> r >> c;

    l--; r--;

    to[l].emplace_back(r,c);

  }

  REP(i,N-1)

    to[i+1].emplace_back(i,0);



  REP(i,N)

    dist[i] =INF;

  dist[0] = 0;

  // ダイクストラ

  priority_queue< pair<ll, int>, vector< pair<ll,int> >, greater<pair<ll,int> > > q;

  q.push( make_pair(0, 0) );

  while( !q.empty() ) {

    auto v = q.top(); q.pop();

    if ( v.first > dist[v.second] ) continue;

    for ( auto nv: to[v.second] ) {

      if ( dist[nv.first] > dist[v.second] + nv.second ) {

        dist[nv.first] = dist[v.second] + nv.second;

        q.push( make_pair(dist[nv.first], nv.first) );

      }

    }

  }

  ll ans = (dist[N-1] == INF) ? -1 : dist[N-1];

  cout << ans << endl;

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<(n);++i)

using namespace std;

typedef long long ll;



const ll INF = 1LL << 60;



ll dp[100005];



// セグメントツリー

//  範囲内の最小値を返す

template <typename T = int>

struct SegmentTree {

  int N;

  vector<T> node;

  T ival;

  SegmentTree( int size, T iv = 0 ) {

    init(size, iv);

  }

  void init( int size, T iv = 0 ) {

    N = 1; while( N < size ) N<<=1;

    node.resize( 2*N-1 );

    ival = iv;

    for(int i=0; i<2*N-1; ++i ) node[i] = ival;

  }

  void update( int pos, T dat ) {

    pos = N + pos - 1;

    node[pos] = dat;

    while ( pos > 0 ) {

      pos = (pos-1)/2;

      node[pos] = min( node[pos*2+1], node[pos*2+2] );

    }

  }

  // [a,b) 区間から取得。現在位置 k, 現在位置の範囲[l,r)

  T get( int a, int b, int k, int l, int r ) {

    if (a>=r || b<=l) return ival;

    if (a<=l && b>=r) return node[k];

    T vl = get( a, b, 2*k+1, l, (l+r)/2 );

//    cout << " vl : " << a << " " <<  b  << " " << 2*k+1 << " " << l << " " << (l+r)/2<< " -> " << vl << endl;

    T vr = get( a, b, 2*k+2, (l+r)/2, r );

//    cout << " vr : " << a << " " <<  b  << " " << 2*k+2 << " " << (l+r)/2 << " " << r << " -> " << vr << endl;

    return min(vl,vr);

  }

  void print() {

    cout << "---- node " << endl;

    for ( int i=0; i<2*N-1; ++i ) {

      cout << " " << i << " : " << node[i] << endl;

    }

  }

  int getRMax() { return N; }

};



int main() {

  int N,M;

  cin >> N >> M;

  vector< tuple<int,int,int> > q;

  SegmentTree<ll> st(N,INF);

  REP(i,M) {

    int l,r,c;

    cin >> l >> r >> c;

    l--; r--;

    q.emplace_back( r, l, c );

  }

  sort( q.begin(), q.end() );

  

  REP(i,N) dp[i] = INF;

  dp[0] = 0;

  

  st.update( 0, dp[0] );

//  st.print();

  REP(i,M) {

    int l,r,c;

    tie(r,l,c) = q[i];

    ll val = st.get(l,r+1,0,0,st.getRMax()) + c;

//    printf("%d %d %d : %lld %lld\n",r,l,c,val,dp[r]);

    if ( dp[r] > val ) {

      dp[r] = val;

      st.update( r, val );

//      st.print();

    }

  }

//  REP(i,N) cout << dp[i] << endl;



  cout << ((dp[N-1]==INF) ? -1 : dp[N-1]) << endl;

}


// Fastest known solution for this problem: