A tourist hiked along the mountain range. The hike lasted for n days, during each day the tourist noted height above the sea level. On the i-th day height was equal to some integer h_i. The tourist pick smooth enough route for his hike, meaning that the between any two consecutive days height changes by at most 1, i.e. for all i's from 1 to n - 1 the inequality |h_i - h_i + 1| ≤ 1 holds.

At the end of the route the tourist rafted down a mountain river and some notes in the journal were washed away. Moreover, the numbers in the notes could have been distorted. Now the tourist wonders what could be the maximum height during his hike. Help him restore the maximum possible value of the maximum height throughout the hike or determine that the notes were so much distorted that they do not represent any possible height values that meet limits |h_i - h_i + 1| ≤ 1.

The first line contains two space-separated numbers, n and m (1 ≤ n ≤ 10⁸, 1 ≤ m ≤ 10⁵) — the number of days of the hike and the number of notes left in the journal.

Next m lines contain two space-separated integers d_i and h_di (1 ≤ d_i ≤ n, 0 ≤ h_di ≤ 10⁸) — the number of the day when the i-th note was made and height on the d_i-th day. It is guaranteed that the notes are given in the chronological order, i.e. for all i from 1 to m - 1 the following condition holds: d_i < d_i + 1.

If the notes aren't contradictory, print a single integer — the maximum possible height value throughout the whole route.

If the notes do not correspond to any set of heights, print a single word 'IMPOSSIBLE' (without the quotes).

For the first sample, an example of a correct height sequence with a maximum of 2: (0, 0, 1, 2, 1, 1, 0, 1).

In the second sample the inequality between h₇ and h₈ does not hold, thus the information is inconsistent.

题意就是一个人爬山，不小心把记录自己爬过山的海拔的数据给(我也不知道怎么搞的)丢了一部分，每次爬山相邻两天海拔距离不超过1，问你爬的最高海拔是多少。。。

eg:

数据（a，b），假设从第一天到第a天一直减1，则第一天的海拔为 b + (a - 1)；从第pre_a天海拔为pre_b，到第a天海拔为b，假设中间的最高海拔为x，则可列出不等式（x - pre_b) + (x - b) <= a - pre_a，整理得x = （pre_b + b + a - pre_a）/ 2；从最后一条数据（a，b）到第n天，假设一直加1，则第n天的海拔为b+(n-a)。每次求出一个值，求出他们的最大值即可。

自己写的代码wa了，然后看题解又写的，菜的抠脚。。。

代码:

#include<bits/stdc++.h>

using namespace std;

int main(){

    int n,m,a,b;

    while(~scanf("%d%d",&n,&m)){

        scanf("%d%d",&a,&b);

        int pre_a=a,pre_b=b;

        int ans=b+a-1;

        bool flag=true;

        for(int i=1;i<m;i++){

            scanf("%d%d",&a,&b);

            if(abs(b-pre_b)>a-pre_a) flag=false;

            int tmp=(pre_b+b+a-pre_a)/2;

            ans=max(ans,tmp);

            pre_a=a;pre_b=b;

        }

        int tmp=pre_b+(n-pre_a);

        ans=max(ans,tmp);

        if(flag) printf("%d\n",ans);

        else printf("IMPOSSIBLE\n");

    }

    return 0;

}