Jamie has recently found undirected weighted graphs with the following properties very interesting:

• The graph is connected and contains exactly n vertices and m edges.
• All edge weights are integers and are in range [1, 10⁹] inclusive.
• The length of shortest path from 1 to n is a prime number.
• The sum of edges' weights in the minimum spanning tree (MST) of the graph is a prime number.
• The graph contains no loops or multi-edges.

If you are not familiar with some terms from the statement you can find definitions of them in notes section.

Help Jamie construct any graph with given number of vertices and edges that is interesting!

Input

First line of input contains 2 integers n, m  — the required number of vertices and edges.

Output

In the first line output 2 integers sp, mstw (1 ≤ sp, mstw ≤ 10¹⁴) — the length of the shortest path and the sum of edges' weights in the minimum spanning tree.

In the next m lines output the edges of the graph. In each line output 3 integers u, v, w (1 ≤ u, v ≤ n, 1 ≤ w ≤ 10⁹) describing the edge connecting u and v and having weight w.

Example

Input

4 4

Output

7 7 1 2 3 2 3 2 3 4 2 2 4 4

Input

5 4

Output

7 13 1 2 2 1 3 4 1 4 3 4 5 4

Note

The graph of sample 1: Shortest path sequence: {1, 2, 3, 4}. MST edges are marked with an asterisk (*).

Definition of terms used in the problem statement:

A shortest path in an undirected graph is a sequence of vertices (v₁, v₂, ... , v_k) such that v_i is adjacent to v_i + 11 ≤ i < k and the sum of weight is minimized where w(i, j) is the edge weight between i and j. ()

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. ()

A minimum spanning tree (MST) is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. ()

 

题意 ： 让你构造一个图，保证图的MST是一个素数，并且1 到 n的路径也是一个素数，输出图的全部边

题目分析 ： 两种构造方法，一是从1到n先构成一条链，让边上的总和是一个素数且等于 1e5+3, 可以让1到n -2 条边全是 1，让最后一条边是素数减去前面的总和，然后再有边的话，我们就给它赋值一个很大的数即可

代码示例 ：

const int prime = 1e5+3;const int maxn = 0x3f3f3f3f;#define ll long longint main() {    int n, m;        cin >> n >> m;    printf("%d %d\n", prime, prime);    for(int i = 1; i < n-1; i++){        printf("%d %d %d\n", i, i+1, 1);    }    printf("%d %d %d\n", n-1, n, prime-(n-2));        int cnt = n-1, sign = 0;    for(int i = 1; i < n; i++){        for(int j = i+2; j <= n; j++){            if (cnt == m) {sign = 1; break;}            printf("%d %d %d\n", i, j, prime*2);            cnt++;        }        if (sign) break;    }     return 0;}

 

第二种构造方法 ： 由 1 这个点去向其余所有的点引边，并别将 1 到 n的路径赋值为 2 ，在有边的话将其余的边赋值为很大的数即可

代码示例 ：

const int prime = 1e5+3;const int maxn = 0x3f3f3f3f;#define ll long longint main() {    int n, m;        cin >> n >> m;    if (n == 2) printf("%d %d\n", 2, 2);    else printf("%d %d\n", 2, prime);    for(int i = 2; i < n-1; i++){        printf("%d %d %d\n", 1, i, 1);    }    if (m > 1)        printf("%d %d %d\n", 1, n-1, prime-2-(n-3));    printf("%d %d %d\n", 1, n, 2);    int cnt = n-1, sign = 0;    for(int i = 2; i < n; i++){        for(int j = i+1; j <= n; j++){            if (cnt == m){sign = 1; break;}            printf("%d %d %d\n", i, j, prime*2);            cnt++;        }        if (sign) break;    }    return 0;}