N children are sitting in a circle to play a game.

The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her hand. The game starts from the K-th child, who tells all the others the integer on his card and jumps out of the circle. The integer on his card tells the next child to jump out. Let A

 denote the integer. If Ais positive, the next child will be the A-th child to the left. If A is negative, the next child will be the (−A)-th child to the right.

The game lasts until all children have jumped out of the circle. During the game, the p-th child jumping out will get F

(p) candies where F(p) is the number of positive integers that perfectly divide p. Who gets the most candies?

Input

There are several test cases in the input. Each test case starts with two integers N(0 < N ≤ 500,000) and K (1 ≤ K ≤ N) on the first line. The next N

 lines contains the names of the children (consisting of at most 10 letters) and the integers (non-zero with magnitudes within 108) on their cards in increasing order of the children’s numbers, a name and an integer separated by a single space in a line with no leading or trailing spaces.

Output

Output one line for each test case containing the name of the luckiest child and the number of candies he/she gets. If ties occur, always choose the child who jumps out of the circle first.

Sample Input

4 2
Tom 2
Jack 4
Mary -1
Sam 1

Sample Output

Sam 3

題解：先把因子最多的那個數找出來，然後模擬進行即可，人數每次減1，線段樹單點查詢

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define lowbit(x) (x&(-x))
const int N=500100;
typedef long long ll;
struct node{
	int l,r;
	int val;
}tree[N<<2];
struct node1{
	char name[15];
	int id;
}a[N];
int n,k,num[N];
void init()
{
	for(int i=1;i<N;i++)
	{
		num[i]++;
		for(int j=i+i;j<N;j+=i)
			num[j]++;
	}
}
int solve()
{
	int ans=0,p;
	for(int i=1;i<=n;i++)
	{
		if(num[i]>ans)
			ans=num[i],p=i;
	}
	return p;
}
void build(int l,int r,int cur)
{
	tree[cur].l=l;
	tree[cur].r=r;
	tree[cur].val=r-l+1;
	if(l==r) return;
	int mid=(r+l)>>1;
	build(l,mid,cur<<1);
	build(mid+1,r,cur<<1|1);
}
int update(int pos,int cur)
{
	tree[cur].val--;
	if(tree[cur].l==tree[cur].r)
	{
		return tree[cur].l;
	}
	if(pos<=tree[cur<<1].val) return update(pos,cur<<1);
	else return update(pos-tree[cur<<1].val,cur<<1|1); 
}
int main()
{
	init();
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=1;i<=n;i++)scanf("%s%d",a[i].name,&a[i].id);
		build(1,n,1);
		int ans=solve(),pos,cnt=k;
		k=update(k,1);
		n--;
		for(int i=2;i<=ans;i++)
		{
			if(a[k].id>0)
			{
				pos=((a[k].id+cnt-1)%n+n)%n;
				if(pos==0) pos=n;
			} 
			if(a[k].id<0)
			{
				pos=((cnt+a[k].id)%n+n)%n;
				if(pos==0) pos=n;
			} 
			
			k=update(pos,1);
			cnt=pos;
			n--;
		}
		printf("%s %d\n",a[k].name,num[ans]);
    }
	return 0;
}