Codeforces Round #434 (Div. 2, based on Technocup 2018 Elimination Round 1)&&Codeforces 861B Which floor?【枚举，暴力】

B. Which floor?

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1.

Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers.

Given this information, is it possible to restore the exact floor for flat n?

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory.

m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers k_i, f_i (1 ≤ k_i ≤ 100, 1 ≤ f_i ≤ 100), which means that the flat k_i is on the f_i-th floor. All values k_i are distinct.

It is guaranteed that the given information is not self-contradictory.

Output

Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor.

Examples

Input

10 3
6 2
2 1
7 3

Output

4

Input

8 4
3 1
6 2
5 2
2 1

Output

-1

Note

In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor.

In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat.

题目链接：http://codeforces.com/contest/861/problem/B

分析：直接看代码吧，代码给出了详细注释！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N=200;
int n,m,cur=0,idx=0;;
int should[N+10];
inline void ok(int per)
{
    int now=1,cnt=0;//now是当前的楼层,cnt是当前楼层的公寓数目
    int temp=1;
    for(int dd=1;dd<=100;dd++)//枚举第dd间房子在哪里
    {
        cnt++;//当前楼层的公寓数目递增。
        if(cnt>per)//如果公寓数目大于每层的楼层数
        {
            cnt=1;//进入下一层
            now++;//楼层个数递增
        }
        if(should[dd]!=0&&should[dd]!=now)
            return;
        if(dd == n)
            temp=now;
        //如果dd公寓不应该在第now层,就结束
    }
    //是一个合法的分配
    if(cur==0)//如果之前没有找到过合法的。
    {
        idx=temp;//第n个房子,它就在第now层
        cur=1;//找到的解数目为1
    }
    else//数目大于0了 
    {
        if(cur==1)//如果只有一个解的话
        {
            if(idx==temp)//如果第n间房子的层数和当前一样,退出
                return;
        }
        cur++;//否则,答案递增。
    }
}
int main(void)
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        should[x]=y;//x号公寓房子啊第y层
    }
    for(int i=1;i<=102;i++)//枚举每层有i间公寓
        ok(i);
    if(cur>1||cur==0)//如果解的个数太多,或没解。
        cout<<-1;
    else
        cout<<idx;
    return 0;
}