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,n) for (int i = 0; i < (n); ++i)
#define all(v) (v).begin(),(v).end()
#define chmin(x,y) x = min(x,y)
typedef pair<int, int> P;
typedef long long ll;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;
const ll MOD = 1e9+7;

int main() {
	int N, M;
	cin >> N >> M;

    vector<tuple<int, int, ll>> LRC(M);
    int l, r;
    ll c;
    rep(i, M) {
        cin >> l >> r >> c;
        --l, --r;
        LRC[i] = forward_as_tuple(l, r, c);
    }
    sort(all(LRC));

    vector<ll> dp(N, LINF);
    dp[0] = 0;

    rep(i, M) {
        tie(l, r, c) = LRC[i];
        // cout << l << r << c << endl;
        if (dp[l] == LINF) continue;
        for (int j = r; j > l; j--) {
            if (dp[j] > dp[l] + c) dp[j] = dp[l] + c;
            else break;
        }       
    }

    // rep(j, N) cout << dp[j] << ' ';

    if (dp[N-1] == LINF) cout << -1 << endl;
    else cout << dp[N-1] << endl;
	return 0;
}