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 N 100005

int L[N], R[N], C[N];

int n, m;

using ll = long long;
ll a[N];

vector <int> v[N];

const ll inf = 0x3f3f3f3f3f3f3f3f;

void give(int k, ll v) {
    k = n + 1 -  k;
    a[k] = v;
}
ll calc(int k) {
	k = n + 1 - k;
    ll res = inf;
    for ( ; k; k -= k & -k) res = min(res, a[k]);
    return res;
}
void update(int k, ll val) {
	k = n + 1 - k;
	for ( ; k <= n; k += k & -k) a[k] = min(a[k], val);
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= m; i ++) scanf("%d %d %d", &L[i], &R[i], &C[i]), v[L[i]].push_back(i);
    for (int i = 1; i <= n; i ++) give(i, inf);
    give(1, 0);
    for (int i = 1; i <= n; i ++) {
        ll val = calc(i);
        for (auto x : v[i]) {
            int r = R[x], c = C[x];
            update(r, val + c);
        }
    }
    ll ans = calc(n);
    if (ans == inf) ans = -1;
    cout << ans << endl;
	return 0;
}