D. Relatively Prime Graph

Let's call an undirected graph $\mathrm{G=(V,E)G=(V,E) relatively\; prime if\; and\; only\; if\; for\; each\; edge (v,u)\in E(v,u)\in E GCD(v,u)=1GCD(v,u)=1 (the\; greatest\; common\; divisor\; of vv and uu is 11).\; If\; there\; is\; no\; edge\; between\; some\; pair\; of\; vertices vv and uu then\; the\; value\; of GCD(v,u)GCD(v,u) doesn\text{'}t\; matter.\; The\; vertices\; are\; numbered\; from 11 to |V||V|.}$

Construct a relatively prime graph with $\mathrm{nn vertices\; and mm edges\; such\; that\; it\; is\; connected\; and\; it\; contains\; neither\; self-loops\; nor\; multiple\; edges.}$

If there exists no valid graph with the given number of vertices and edges then output "Impossible".

If there are multiple answers then print any of them.

Input

The only line contains two integers $\mathrm{nn and mm (1\le n,m\le 1051\le n,m\le 105)\; —\; the\; number\; of\; vertices\; and\; the\; number\; of\; edges.}$

Output

If there exists no valid graph with the given number of vertices and edges then output "Impossible".

Otherwise print the answer in the following format:

The first line should contain the word "Possible".

The $\mathrm{ii-th\; of\; the\; next mm lines\; should\; contain\; the ii-th\; edge (vi,ui)(vi,ui) of\; the\; resulting\; graph\; (1\le vi,ui\le n,vi\ne ui1\le vi,ui\le n,vi\ne ui).\; For\; each\; pair (v,u)(v,u)there\; can\; be\; no\; more\; pairs (v,u)(v,u) or (u,v)(u,v).\; The\; vertices\; are\; numbered\; from 11 to nn.}$

If there are multiple answers then print any of them.

Examples

input

Copy

5 6

output

Copy

Possible
2 5
3 2
5 1
3 4
4 1
5 4

input

Copy

6 12

output

Copy

Impossible

Note

Here is the representation of the graph from the first example:

   这题无脑暴力 暴力真的出了奇迹 

   暴力枚举一遍就行了

    

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int INF = 0x3fffffff;
typedef long long LL;
using namespace std;
int n, m;
struct node {
    int x, y;
    node () {}
    node (int x, int y): x(x), y(y) {}
} qu[maxn];
int main() {
    scanf("%d%d", &n, &m);
    if (n - 1 > m) {
        printf("Impossible
");
        return 0;
    }
    int k = 0, flag = 0;
    for (int i = 1 ; i <= n ; i++) {
        for (int j = i + 1 ; j <= n ; j++) {
            if (__gcd(i, j) == 1) qu[k++] = node(i, j);
            if (k == m) {
                flag = 1;
                break;
            }
        }
        if (flag) break;
    }
    if (flag) {
        printf("Possible
");
        for (int i = 0 ; i < k ; i++)
            printf("%d %d
", qu[i].x, qu[i].y);
    } else  printf("Impossible
");
    return 0;
}