CodeForces 589F

F. Gourmet and Banquet

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

A gourmet came into the banquet hall, where the cooks suggestedn dishes for guests. The gourmet knows the schedule: when each of the dishes will be served.

For i-th of the dishes he knows two integer moments in timea_i andb_i (in seconds from the beginning of the banquet) — when the cooks will bring thei-th dish into the hall and when they will carry it out (a_i < b_i). For example, ifa_i = 10 andb_i = 11, then thei-th dish is available for eating during one second.

The dishes come in very large quantities, so it is guaranteed that as long as the dish is available for eating (i. e. while it is in the hall) it cannot run out.

The gourmet wants to try each of the n dishes and not to offend any of the cooks. Because of that the gourmet wants to eat each of the dishes forthe same amount of time. During eating the gourmet can instantly switch between the dishes. Switching between dishes is allowed for him only at integer moments in time. The gourmet can eat no more than one dish simultaneously. It is allowed to return to a dish after eating any other dishes.

The gourmet wants to eat as long as possible on the banquet without violating any conditions described above. Can you help him and find out the maximum total time he can eat the dishes on the banquet?

Input

The first line of input contains an integer n (1 ≤ n ≤ 100) — the number of dishes on the banquet.

The following n lines contain information about availability of the dishes. Thei-th line contains two integers a_i and b_i (0 ≤ a_i < b_i ≤ 10000) — the moments in time when thei-th dish becomes available for eating and when thei-th dish is taken away from the hall.

Output

Output should contain the only integer — the maximum total time the gourmet can eat the dishes on the banquet.

The gourmet can instantly switch between the dishes but only at integer moments in time. It is allowed to return to a dish after eating any other dishes. Also in every moment in time he can eat no more than one dish.

Examples

Input

32 41 56 9

Output

6

Input

31 21 21 2

Output

0

Note

In the first example the gourmet eats the second dish for one second (from the moment in time 1 to the moment in time 2), then he eats the first dish for two seconds (from 2 to 4), then he returns to the second dish for one second (from 4 to 5). After that he eats the third dish for two seconds (from 6 to 8).

In the second example the gourmet cannot eat each dish for at least one second because there are three dishes but they are available for only one second (from 1 to 2).

n道菜，每道菜吃相同的时间，求总时间的最大值。

将每个区间的右端点排序，使用贪心思想，保证最优。

从0到区间最大长度开始二分。

#include<cstdio>#include<string.h>#include<algorithm>using namespace std;int n;int ans;int vis[10005];struct node{    int a,b;    bool operator < (const node& tmp) const    {        return b<tmp.b||(b==tmp.b&&a<tmp.a);    }}dish[105];bool fun(int m){    memset(vis,0,sizeof(vis));    int i;    for(i=0;i<n;i++)    {        if(dish[i].b-dish[i].a<m)  return false;        int cur=0;        int j;        for(j=dish[i].a;j<dish[i].b;j++)        {            if(!vis[j])            {                cur++;                vis[j]=1;                if(cur==m) break;            }        }        if(cur<m) return false;    }    return true;}int main(){    while(scanf("%d",&n)!=EOF)    {        int i;        int right=0,left=0;        ans=0;        for(i=0;i<n;i++)        {            scanf("%d%d",&dish[i].a,&dish[i].b);            if(dish[i].b-dish[i].a>right)                right=dish[i].b-dish[i].a;        }        sort(dish,dish+n);        while(left<=right)        {            int mid=(left+right)>>1;            if(fun(mid))            {                left=mid+1;                ans=max(ans,mid);            }            else                right=mid-1;        }        printf("%d\n",ans*n);    }    return 0;}