Round #432 (Div.2)

A. Arpa and a research in Mexican wave

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Arpa is researching the Mexican wave.

There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.

• At time 1, the first spectator stands.
• At time 2, the second spectator stands.
• ...
• At time k, the k-th spectator stands.
• At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
• At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
• ...
• At time n, the n-th spectator stands and the (n - k)-th spectator sits.
• At time n + 1, the (n + 1 - k)-th spectator sits.
• ...
• At time n + k, the n-th spectator sits.

Arpa wants to know how many spectators are standing at time t.

 

Input

The first line contains three integers n, k, t (1 ≤ n ≤ 10⁹, 1 ≤ k ≤ n, 1 ≤ t < n + k).

 

Output

Print single integer: how many spectators are standing at time t.

 

Examples

 

Input

10 5 3

 

Output

3

 

Input

10 5 7

 

Output

5

 

Input

10 5 12

 

Output

3

 

Note

In the following a sitting spectator is represented as -, a standing spectator is represented as ^.

• At t = 0  ---------- number of standing spectators = 0.
• At t = 1  ^--------- number of standing spectators = 1.
• At t = 2  ^^-------- number of standing spectators = 2.
• At t = 3  ^^^------- number of standing spectators = 3.
• At t = 4  ^^^^------ number of standing spectators = 4.
• At t = 5  ^^^^^----- number of standing spectators = 5.
• At t = 6  -^^^^^---- number of standing spectators = 5.
• At t = 7  --^^^^^--- number of standing spectators = 5.
• At t = 8  ---^^^^^-- number of standing spectators = 5.
• At t = 9  ----^^^^^- number of standing spectators = 5.
• At t = 10 -----^^^^^ number of standing spectators = 5.
• At t = 11 ------^^^^ number of standing spectators = 4.
• At t = 12 -------^^^ number of standing spectators = 3.
• At t = 13 --------^^ number of standing spectators = 2.
• At t = 14 ---------^ number of standing spectators = 1.
• At t = 15 ---------- number of standing spectators = 0.

#include <iostream> 
#include <string.h> 
using namespace std;
int main(){
    int n,k,t;
    while(cin>>n>>k>>t){
        if(t<k) cout<<t<<endl;
        else if(t>n) cout<<k-(t-n)<<endl;
        else cout<<k<<endl;
    }
    
}

B. Arpa and an exam about geometry

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

 

Input

The only line contains six integers a_x, a_y, b_x, b_y, c_x, c_y (|a_x|, |a_y|, |b_x|, |b_y|, |c_x|, |c_y| ≤ 10⁹). It's guaranteed that the points are distinct.

 

Output

Print "Yes" if the problem has a solution, "No" otherwise.

You can print each letter in any case (upper or lower).

 

Examples

 

Input

0 1 1 1 1 0

 

Output

Yes

 

Input

1 1 0 0 1000 1000

 

Output

No

Note

In the first sample test, rotate the page around (0.5, 0.5) by .

In the second sample test, you can't find any solution.

#include <bits/stdc++.h>
using namespace std;
int main(){
    double ax,ay,bx,by,cx,cy;
    while(cin>>ax>>ay>>bx>>by>>cx>>cy){
        double d1=(by-ay)*(by-ay)+(bx-ax)*(bx-ax);
        double d2=(cy-by)*(cy-by)+(cx-bx)*(cx-bx);
        if(d1!=d2) cout<<"No"<<endl;
        else {
            double k1=(by-ay)/(bx-ax);
            double k2=(cy-by)/(cx-bx);
            if(k1==k2) cout<<"No"<<endl;
            else cout<<"Yes"<<endl;
        } 
    } 
    
}