High Load CodeForces - 828D

Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of n nodes connected with minimum possible number of wires into one network (a wire directly connects two nodes). Exactly k of the nodes should be exit-nodes, that means that each of them should be connected to exactly one other node of the network, while all other nodes should be connected to at least two nodes in order to increase the system stability.

Arkady wants to make the system as fast as possible, so he wants to minimize the maximum distance between two exit-nodes. The distance between two nodes is the number of wires a package needs to go through between those two nodes.

Help Arkady to find such a way to build the network that the distance between the two most distant exit-nodes is as small as possible.

Input

The first line contains two integers n and k (3 ≤ n ≤ 2·10⁵, 2 ≤ k ≤ n - 1) — the total number of nodes and the number of exit-nodes.

Note that it is always possible to build at least one network with nnodes and k exit-nodes within the given constraints.

Output

In the first line print the minimum possible distance between the two most distant exit-nodes. In each of the next n - 1 lines print two integers: the ids of the nodes connected by a wire. The description of each wire should be printed exactly once. You can print wires and wires' ends in arbitrary order. The nodes should be numbered from 1to n. Exit-nodes can have any ids.

If there are multiple answers, print any of them.

Example

Input

3 2

Output

2
1 2
2 3

Input

5 3

Output

3
1 2
2 3
3 4
3 5

Note

In the first example the only network is shown on the left picture.

In the second example one of optimal networks is shown on the right picture.

Exit-nodes are highlighted.

该题解法挺多，我对题目所需要的最小值进行二分。判断该二分值时，已知二分假设的最短距离以及总点数。可以按照这个假设值构成一棵树的最长边，在该边中心处连点，求出所需要的最少叶子数。若该最小叶子数小于等于k，那么该假设成立可以继续缩小，否则增大；最后找到边界值。

#include"stdio.h"
#include"algorithm"
#include"string.h"
#include"cstdio"
using namespace std;
const int max_n=2*(1e5)+10;
int n,k;
int len(int mid)
{
    int maxx=mid/2;
    if(maxx==0)
    return k;
    int less=(n-mid-1)%maxx==0?(n-mid-1)/maxx:(n-mid-1)/maxx+1;
    return less;
}
int EF()
{
    int l=2,r=n-1,mid;
    mid=(l+r)>>1;
    while(l<=r)
    {
        int maxx=mid/2;
        int less=len(mid);
        if(less+2<=k)
        r=mid-1;
        else 
        l=mid+1;
        mid=(l+r)>>1;//printf("&&  %d %d mid:%d\n",l,r,mid);//printf("**\n");
    }
    if(len(r)+2<=k)
    return r;
    return l;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
         int ret=EF();
         printf("%d\n",ret);
        int i,num=0;
        int point[max_n],root=0;
        for(i=1;i<=ret;i++)
        printf("%d %d\n",i,i+1);
        i++;
        for(int j=0;j<k-2;j++)
            {
                printf("%d %d\n",(ret+2)/2,i);
                point[++root]=i++;
            }
        while(i<=n)
        {
            
            printf("%d %d\n",point[root--],i++);
            for(int p=0;p<(ret/2-2)&&i<=n;)
            {
                printf("%d %d\n",i-1,i);
                i++; p++;
            }
        }
    }
}