Codeforces Round #392 (Div. 2) C - Unfair Poll codeforces

C. Unfair Poll

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others.

Seating in the class looks like a rectangle, where n rows with m pupils in each. 

The teacher asks pupils in the following order: at first, she asks all pupils from the first row in the order of their seating, then she continues to ask pupils from the next row. If the teacher asked the last row, then the direction of the poll changes, it means that she asks the previous row. The order of asking the rows looks as follows: the 1-st row, the 2-nd row, ..., the n - 1-st row, the n-th row, the n - 1-st row, ..., the 2-nd row, the 1-st row, the 2-nd row, ...

The order of asking of pupils on the same row is always the same: the 1-st pupil, the 2-nd pupil, ..., the m-th pupil.

During the lesson the teacher managed to ask exactly k questions from pupils in order described above. Sergei seats on the x-th row, on the y-th place in the row. Sergei decided to prove to the teacher that pupils are asked irregularly, help him count three values:

1. the maximum number of questions a particular pupil is asked, 
2. the minimum number of questions a particular pupil is asked, 
3. how many times the teacher asked Sergei. 

If there is only one row in the class, then the teacher always asks children from this row.

Input

The first and the only line contains five integers n, m, k, x and y (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018, 1 ≤ x ≤ n, 1 ≤ y ≤ m).

Output

Print three integers:

1. the maximum number of questions a particular pupil is asked, 
2. the minimum number of questions a particular pupil is asked, 
3. how many times the teacher asked Sergei. 

Examples

input

1 3 8 1 1

output

3 2 3

input

4 2 9 4 2

output

2 1 1

input

5 5 25 4 3

output

1 1 1

input

100 100 1000000000000000000 100 100

output

101010101010101 50505050505051 50505050505051

Note

The order of asking pupils in the first test: 

1. the pupil from the first row who seats at the first table, it means it is Sergei; 
2. the pupil from the first row who seats at the second table; 
3. the pupil from the first row who seats at the third table; 
4. the pupil from the first row who seats at the first table, it means it is Sergei; 
5. the pupil from the first row who seats at the second table; 
6. the pupil from the first row who seats at the third table; 
7. the pupil from the first row who seats at the first table, it means it is Sergei; 
8. the pupil from the first row who seats at the second table; 

The order of asking pupils in the second test: 

1. the pupil from the first row who seats at the first table; 
2. the pupil from the first row who seats at the second table; 
3. the pupil from the second row who seats at the first table; 
4. the pupil from the second row who seats at the second table; 
5. the pupil from the third row who seats at the first table; 
6. the pupil from the third row who seats at the second table; 
7. the pupil from the fourth row who seats at the first table; 
8. the pupil from the fourth row who seats at the second table, it means it is Sergei; 
9. the pupil from the third row who seats at the first table;

题目大意：有一个n*m大小的教室，每个座位都坐上了人，老师按照从第一排到最后一排，在从倒数第二排到第一排，再从第一排到最后一排，每排从第一个人到第m个人，这么提问k次，问提问最多的人的提问次数是多少，最少的提问次数，和x，y坐标的那个座位的提问次数。

像图片那样，第一行到最后一行，在到第二行为一个循环，每个循环第一行和最后一行到递增为1，中间那些递增2，所以我们得出2*(n-1)*m个格子为一个周期，k／t*2 为中间的增长整周期。k／t 为头尾增长整周期， k%t 为整周期之后剩余的提问次数。

#include "stdio.h"
#include "string.h"
#include "algorithm"
using namespace std;
typedef long long ll;
int n,m,X,Y;
ll K,ans[110][110];
int main()
{
    int i,j,t;
    long long mn = 1e18, mx = 0;
    scanf("%d%d%lld%d%d",&n,&m,&K,&X,&Y);
    t = (n > 1 ? (2*n-2)*m : m);
    for(i=2;i<n;i++)
        for(j=1;j<=m;j++)
            ans[i][j] = K/t*2;
    for(j=1;j<=m;j++)
        ans[1][j] = ans[n][j] = K/t;
    K %= t;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            if(K > 0)
                ans[i][j]++, K--;
    for(i=n-1;i>1;i--)
        for(j=1;j<=m;j++)
            if(K > 0)
                ans[i][j]++, K--;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++){
            mn = (mn < ans[i][j] ? mn : ans[i][j]);
            mx = (mx > ans[i][j] ? mx : ans[i][j]);
        }
    printf("%lld %lld %lld\n",mx,mn,ans[X][Y]);
    return 0;
}