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;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef unsigned long long ll;
typedef long double ld;
typedef pair<int,int> P;
constexpr ll INF = (1LL << 60);

int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<P>> g(n);
  for (int i = 1; i < n; i++) {
    g[i].emplace_back(make_pair(i - 1, 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<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
  vector<ll> dp(n, INF);
  dp[0] = 0;
  pq.push(make_pair(0LL, 0));

  while (!pq.empty()) {
    auto p = pq.top(); pq.pop();
    int u = p.second;
    if (dp[u] < p.first) continue;
    for (auto e : g[u]) {
      if (dp[e.first] > dp[u] + e.second) {
        dp[e.first] = dp[u] + e.second;
        pq.push(make_pair(dp[e.first], e.first));
      }
    }
  }
  ll res = dp[n-1];
  if (res == INF) cout << -1 << endl;
  else cout << res << endl;
  return 0;
}