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>
#define mmp(a,b) make_pair(a,b)
using namespace std;
typedef long long ll;
const int MN = 100100;
const ll INFL = 1LL<<61;
typedef vector<ll> vl;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
bitset<MN> bs;
vpl st[MN];
vl ed[MN];
typedef vector<pl> vpl;
ll res[MN];
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	priority_queue<pl,vector<pl>,greater<pl> > pq;
	ll n,m;
	cin >> n >> m;
	fill(res,res+n,INFL);
	for(int i=0;i<m;i++) {
		ll a,b,c;
		cin >> a >> b >> c;
		a--;b--;
		st[a].push_back(mmp(i,c));
		ed[b+1].push_back(i);
	}
	pq.push({INFL,m});
	res[0] = 0;
	for(int i=0;i<n;i++) {
		for(int j=0;j<ed[i].size();j++) {
			bs.set(ed[i][j]);
		}
		while(bs[pq.top().second]) {
			pq.pop();
		}
		res[i] = min(res[i],pq.top().first);
		for(int j=0;j<st[i].size();j++) {
			ll id = st[i][j].first,co = st[i][j].second;
			ll psh = min(co+res[i],INFL);
			pq.push(mmp(psh,id));
		}
	}
	if(res[n-1] == INFL) {
		cout << -1 << '\n';
	} else {
		cout << res[n-1] << '\n';
	}
}