Problem: 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.

Examples

Input

4 3
1 3 2
2 4 3
1 4 6


Output

5


Input

4 2
1 2 1
3 4 2


Output

-1


Input

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


Output

28

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

const int64_t INF = INT64_C(1) << 60;

template <typename T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  vector<vector<pair<int, int>>> G(N);
  for (int i = 0; i < N - 1; i++) {
    G[i + 1].emplace_back(make_pair(i, 0));
  }
  for (int i = 0; i < M; i++) {
    int l, r, c;
    cin >> l >> r >> c;
    l--, r--;
    G[l].emplace_back(make_pair(r, c));
  }
  priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>> que;
  que.push(make_pair(0, 0));
  vector<int64_t> dist(N, INF);
  dist[0] = 0;
  while (!que.empty()) {
    auto q = que.top();
    que.pop();
    int u = q.second;
    for (auto g : G[u]) {
      int v = g.first, c = g.second;
      if (chmin(dist[v], dist[u] + c)) {
        que.push(make_pair(dist[v], v));
      }
    }
  }
  if (dist[N - 1] != INF) {
    cout << dist[N - 1] << '\n';
  } else {
    cout << -1 << '\n';
  }
  return 0;
}