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, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

int n, m;
ll ans = -1;

int main() {
	cin.tie(0)->sync_with_stdio(0);
    cin >> n >> m;
    vector<pair<pii, int>> edges;
    for (int i = 0; i < m; ++i) {
        int l, r, c;
        cin >> l >> r >> c;
        edges.push_back({{l, r}, c});
    }

    sort(all(edges));

    // Dist, To which node
    set<pair<ll, ll>> cur;
    cur.insert({0, 1});
    
    for (int i = 0; i < m; ++i) {
        int l, r, c;
        l = edges[i].first.first, r = edges[i].first.second, c = edges[i].second;
        while (sz(cur) && cur.begin()->second < l)
            cur.erase(cur.begin());

        if (!cur.size()) break;
        if (r == n) {
            if (ans == -1) ans = cur.begin()->first + c;
            else ans = min(ans, cur.begin()->first + c);
        }

        cur.insert({cur.begin()->first + c, r});
    }

    cout << ans;
}