【CodeForces - 758C Unfair Poll】 暴力 + 模拟

D - Unfair Poll

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.

Example

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;

题意：一个教室中有n排m列的座位，现在老师会进行k次询问，从第一排第一列到最后一排最后一列，再从最后一排第一列到第一排最后一列（最后一排和第一排只询问了一次，每排从第一列开始询问），问最大的询问次数，最小的询问次数和一个特殊位置的询问次数。

分析：首先我们找到一个循环节，就是从第一排到最后一排，再回到第二排，这就是一次循环。有很多细节，计算循环节的时候不能直接用m*(2*n-2)，因为n的值可能为1。设数组a[i][j]为对应座位的询问次数，因为第一排和最后一排只询问一次，这个时候就要特判n是否等于1，避免第一排和最后一排的a[i][j]赋值时重复。

代码如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;
const int MX = 105;
const int mod = 1e9 + 7;
const LL INF = 1e18 + 5;

LL a[MX][MX];

int main(){
    LL n, m, k, x, y;
    scanf("%I64d%I64d%I64d%I64d%I64d", &n, &m, &k, &x, &y);
    int sum = 0;
    for(int i = 1; i <= n; i++){
        sum += m;
    }
    for(int i = n-1; i >= 2; i--){
        sum += m;
    }
    LL p = k / sum;
    for(int i = 2; i <= n-1; i++){
        for(int j = 1; j <= m; j++){
            a[i][j] += 2*p;
        }
    }
    for(int j = 1; j <= m; j++){
        a[1][j] += p;
    }
    if(n != 1){
        for(int j = 1; j <= m; j++){
            a[n][j] += p;
        }
    }
    k -= sum * p;
    if(k){
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                a[i][j] += 1;
                k--;
                if(!k)  break;
            }
            if(!k)  break;
        }
    }
    if(k){
        for(int i = n-1; i >= 2; i--){
            for(int j = 1; j <= m; j++){
                a[i][j] += 1;
                k--;
                if(!k)  break;
            }
            if(!k)  break;
        }
    }
    LL ans1 = 0, ans2 = INF;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            ans1 = max(ans1, a[i][j]);
            ans2 = min(ans2, a[i][j]);
        }
    }
    printf("%I64d %I64d %I64d\n", ans1, ans2, a[x][y]);
    return 0;
}