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;
typedef pair<ll, int> P;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	b += MOD;
	a += b;
	a %= MOD;
}

ll d[100100];
vector<P> g[100100];
priority_queue<P, vector<P>, greater<P> > pq;

int main(void){
	int n,m,l,r,i;
	ll c;
	fill(d,d+100100,INFLL);
	d[0] = 0ll;
	pq.push(P(d[0],0));
	cin >> n >> m;
	for(i=1; i<n; ++i){
		g[i].push_back(P(0ll,i-1));
	}
	for(i=0; i<m; ++i){
		cin >> l >> r >> c;
		--l; --r;
		g[l].push_back(P(c,r));
	}
	while(!pq.empty()){
		l = pq.top().second;
		c = pq.top().first;
		pq.pop();
		if(c>d[l]){
			continue;
		}
		for(i=(g[l].l_ength()-1); i>=0; --i){
			r = g[l][i].second;
			c = g[l][i].first;
			if(d[r] > d[l] + c){
				d[r] = d[l] + c;
				pq.push(P(d[r],r));
			}
		}
	}
	cout << ((d[n-1]<INFLL)?d[n-1]:(-1ll)) << endl;
	return 0;
}