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;
typedef long long ll;

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<pair<int,int>>> q(n);
  for (int i = 0; i < m; i++) {
    int l, r, c;
    cin >> l >> r >> c;
    q[r-1].emplace_back(l-1,c);
  }
  ll B = 1;
  while (B < n) B *= 2;
  vector<ll> mi(B*2, 1e18);
  for (int i = 0; i < n; i++) {
    ll v = 1e18;
    if (i == 0) v = 0;
    for (auto p : q[i]) {
      int l = p.first, c = p.second;
      ll val = mi[l+=B];
      for (; l; l >>= 1) {
	if (l%2 == 0)
	  val = min(val, mi[l+1]);
      }
      v = min(v, val+c);
    }
    mi[i+B] = v;
    for (int j = (i+B)/2; j; j >>= 1) {
      mi[j] = min(mi[j*2], mi[j*2+1]);
    }
  }
  ll ans = mi[B+n-1];
  if (ans >= 1e17) cout << -1 << endl;
  else cout << ans << endl;
}