Problem:
Score : 600 points

Problem Statement

We have N points numbered 1 to N arranged in a line in this order.

Takahashi decides to make an undirected graph, using these points as the vertices.
In the beginning, the graph has no edge. Takahashi will do M operations to add edges in this graph.
The i-th operation is as follows:

The operation uses integers L_i and R_i between 1 and N (inclusive), and a positive integer C_i. For every pair of integers (s, t) such that L_i \leq s < t \leq R_i, add an edge of length C_i between Vertex s and Vertex t.

The integers L_1, ..., L_M, R_1, ..., R_M, C_1, ..., C_M are all given as input.

Takahashi wants to solve the shortest path problem in the final graph obtained. Find the length of the shortest path from Vertex 1 to Vertex N in the final graph.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq L_i < R_i \leq N

1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 C_1
:
L_M R_M C_M

Output

Print the length of the shortest path from Vertex 1 to Vertex N in the final graph.
If there is no shortest path, print -1 instead.

Sample Input 1

4 3
1 3 2
2 4 3
1 4 6

Sample Output 1

5

We have an edge of length 2 between Vertex 1 and Vertex 2, and an edge of length 3 between Vertex 2 and Vertex 4, so there is a path of length 5 between Vertex 1 and Vertex 4.

Sample Input 2

4 2
1 2 1
3 4 2

Sample Output 2

-1

Sample Input 3

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3

Sample Output 3

28

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <vector>

using namespace std;

// BEGIN NO SAD
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
typedef vector<int> vi;
// END NO SAD

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, int> pll;

typedef pair<pii, int> state;

inline bool upd(int k, ll v, map<int, ll>& dp) {
  if(!dp.count(k) || v < dp[k]) {
    dp[k] = v;
    while(true) {
      auto it = dp.lower_bound(k);
      if(it == dp.begin()) break;
      it--;
      if(it->second > dp[k]) {
        dp[it->first] = k;
        k = it->first;
      }
      else break;
    }
    return true;
  }
  return false;
}

void solve() {
  int n, m;
  cin >> n >> m;
  vector<state> v;
  for(int i = 0; i < n; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    v.push_back({pii(a, b), c});
  }
  sort(v.begin(), v.end());
  map<int, ll> dp;
  dp[1] = 0;
  int last = 1;
  int idx = 0;
  while(idx < sz(v)) {
    // cerr << idx << ": " << last << " " << dp[last] << endl;
    if(last < v[idx].first.first) {
      if(dp.upper_bound(last) == dp.end()) break;
      last = dp.upper_bound(last)->first;
      continue;
    }
    if(last >= v[idx].first.second) {
      idx++;
      continue;
    }
    assert(last < v[idx].first.second);
    upd(v[idx].first.second, dp[last] + v[idx].second, dp);
    while(dp.upper_bound(last) != dp.end()) {
      int cand = dp.upper_bound(last)->first;
      if(cand >= v[idx].first.first && cand < v[idx].first.second && upd(v[idx].first.second, dp[cand] + v[idx].second, dp)) {
        last = cand;
      }
      else {
        break;
      }
    }
    idx++;
  }
  if(!dp.count(n)) cout << "-1\n";
  else cout << dp[n] << "\n";
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
}