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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <set>
#include <map>
#include <list>

using namespace std;

typedef long long int ll;
typedef pair<int, int> Pii;

const ll mod = 998244353;
const ll inf = (ll) 1e18;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m;
  cin >> n >> m;
  vector<ll> l(m), r(m), c(m);
  for (int i = 0; i < m; i++) cin >> l[i] >> r[i] >> c[i];

  vector<vector<ll> > lrc(m, vector<ll>(3));
  for (int i = 0; i < m; i++) {
    lrc[i][0] = l[i] - 1;
    lrc[i][1] = -r[i];
    lrc[i][2] = c[i];
  }

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

  for (int i = 0; i < m; i++) lrc[i][1] = -lrc[i][1];

  vector<ll> dp(n+2, inf);
  dp[0] = 0;
  for (int k = 0; k < m; k++) {
    ll b = lrc[k][0];
    ll e = lrc[k][1];
    ll cost = dp[b] + lrc[k][2];
    auto it = upper_bound(dp.begin() + b, dp.begin() + e, cost);
    ll d = distance(dp.begin(), it);
    for (int i = d; i < e; i++) {
      dp[i] = cost;
    }
  }

  auto ans = dp[n-1];
  if (ans >= inf) {
    cout << -1 << endl; // impossible
  }
  else {
    cout << ans << endl;
  }

  return 0;
}