#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pli = pair<ll, int>;
const int N = 1e5 + 5;
const ll INF = 0x3f3f3f3f3f3f3f3f;

ll d[N];
int n, m;
vector<pair<int, ll>> G[N];

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i < n; ++i)
        G[i + 1].emplace_back(i, 0);
    while (m--) {
        int u, v, w;
        cin >> u >> v >> w;
        G[u].emplace_back(v, w);
    }
    memset(d, 0x3f, sizeof d);
    d[1] = 0;
    priority_queue<pli, vector<pli>, greater<pli>> pq;
    pq.emplace(0, 1);
    while (pq.size()) {
        auto p = pq.top(); pq.pop();
        ll cur = p.first;
        int u = p.second;
        if (cur > d[u]) continue;
        for (auto e : G[u]) {
            int v = e.first;
            ll w = e.second;
            if (d[v] > cur + w) {
                d[v] = cur + w;
                pq.emplace(d[v], v);
            }
        }
    }
    cout << (d[n] == INF ? -1 : d[n]) << '\n';
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pli = pair<ll, int>;
const int N = 1e5 + 5;
const ll INF = 0x3f3f3f3f3f3f3f3f;

ll d[N];
int n, m;
vector<pair<int, ll>> G[N];

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i < n; ++i)
        G[i + 1].emplace_back(i, 0);
    while (m--) {
        int u, v, w;
        cin >> u >> v >> w;
        G[u].emplace_back(v, w);
    }
    memset(d, 0x3f, sizeof d);
    d[1] = 0;
    priority_queue<pli, vector<pli>, greater<pli>> pq;
    pq.emplace(0, 1);
    while (pq.size()) {
        auto p = pq.top(); pq.pop();
        ll cur = p.first;
        int u = p.second;
        if (cur > d[u]) continue;
        for (auto e : G[u]) {
            int v = e.first;
            ll w = e.second;
            if (d[v] > cur + w) {
                d[v] = cur + w;
                pq.emplace(d[v], v);
            }
        }
    }
    cout << (d[n] == INF ? -1 : d[n]) << '\n';
}
```