Problem:
Score : 100 points

Problem Statement

You are given an integer N. Consider an infinite N-ary tree as shown below:

Figure: an infinite N-ary tree for the case N = 3

As shown in the figure, each vertex is indexed with a unique positive integer, and for every positive integer there is a vertex indexed with it. The root of the tree has the index 1. For the remaining vertices, vertices in the upper row have smaller indices than those in the lower row. Among the vertices in the same row, a vertex that is more to the left has a smaller index.

Regarding this tree, process Q queries. The i-th query is as follows:

Find the index of the lowest common ancestor (see Notes) of Vertex v_i and w_i.

Notes

In a rooted tree, the lowest common ancestor (LCA) of Vertex v and w is the farthest vertex from the root that is an ancestor of both Vertex v and w. Here, a vertex is considered to be an ancestor of itself. For example, in the tree shown in Problem Statement, the LCA of Vertex 5 and 7 is Vertex 2, the LCA of Vertex 8 and 11 is Vertex 1, and the LCA of Vertex 3 and 9 is Vertex 3.

Constraints

1 ≤ N ≤ 10^9

1 ≤ Q ≤ 10^5

1 ≤ v_i < w_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N Q
v_1 w_1
:
v_Q w_Q

Output

Print Q lines. The i-th line (1 ≤ i ≤ Q) must contain the index of the lowest common ancestor of Vertex v_i and w_i.

Sample Input 1

3 3
5 7
8 11
3 9

Sample Output 1

2
1
3

The queries in this case correspond to the examples shown in Notes.

Sample Input 2

100000 2
1 2
3 4

Sample Output 2

1
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstdlib>
#define cmax(a,b) (a<(b)?a=(b),1:0)
#define cmin(a,b) (a>(b)?a=(b),1:0)
#define dmin(a,b) ((a)<(b)?(a):(b))
#define dmax(a,b) ((a)>(b)?(a):(b))
#define regsiter register
#define CL fclose(stdin),fclose(stdout)
namespace io
{
	int F()
	{
		int n=0,F=1;
		char ch;
		while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		ch=='-'?F=0:n=ch-'0';
		while((ch=getchar())>='0'&&ch<='9')n=(n<<1)+(n<<3)+ch-'0';
		return F?n:-n;
	}
	long long G()
	{
		long long n=0,F=1;
		char ch;
		while((ch=getchar())!='-'&&(ch<'0'||ch>'9'));
		ch=='-'?F=0:n=ch-'0';
		while((ch=getchar())>='0'&&ch<='9')n=(n<<1)+(n<<3)+ch-'0';
		return F?n:-n;
	}
}
int p0[111],p1[111],pp0,pp1;
int main()
{
	int n=io::F(),q=io::F();
	while(q--)
	{
		int u=io::F(),v=io::F();
		if(n==1)printf("%d\n",dmin(u,v));
		pp0=pp1=0;
		for(register int i=u;i;i=(i+n-2)/n)p0[++pp0]=i;
		for(register int i=v;i;i=(i+n-2)/n)p1[++pp1]=i;
		register int x,y;
		for(x=pp0,y=pp1;x&&y&&p0[x]==p1[y];--x,--y);
		printf("%d\n",p0[x+1]);
	}
	return 0;
}