Codeforces Beta Round #91 (Div. 2 Only) D. Lucky Transformation

原题链接

D. Lucky Transformation

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7.
For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are
not.

Petya has a number consisting of n digits without leading zeroes. He represented it as an array of digits without leading
zeroes. Let's call it d. The numeration starts with 1,
starting from the most significant digit. Petya wants to perform the following operation k times:
find the minimum x (1 ≤ x < n) such
that dx = 4 and dx + 1 = 7,
if x is odd, then to assign dx = dx + 1 = 4,
otherwise to assign dx = dx + 1 = 7.
Note that if no x was found, then the operation counts as completed and the array doesn't change at all.

You are given the initial number as an array of digits and the number k. Help Petya find the result of completing k operations.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 109) —
the number of digits in the number and the number of completed operations. The second line contains n digits without spaces representing the array of
digits d, starting with d1.
It is guaranteed that the first digit of the number does not equal zero.

Output

In the single line print the result without spaces — the number after the k operations are fulfilled.

Examples

input

7 4
4727447

output

4427477

input

4 2
4478

output

4478

Note

In the first sample the number changes in the following sequence: 4727447 → 4427447 → 4427477 → 4427447 → 4427477.

In the second sample: 4478 → 4778 → 4478.

447(第一个4为奇数位), 477(4为奇数位)在这两种方式下操作会无限循环

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <cstring>
#define maxn 100005
using namespace std;
typedef long long ll;

int num[maxn];
int main(){

// freopen("in.txt", "r", stdin);
int n, k;
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; i++)
scanf("%1d", num+i);
for(int i = 1; i < n && k; i++){
if(num[i] == 4 && num[i+1] == 7){
if(i % 2 == 0 && num[i-1] == 4){
if(k % 2){
num[i] = 7;
}
break;
}
if(i % 2 == 1 && i + 2 <= n && num[i+2] == 7){
if(k % 2){
num[i+1] = 4;
}
break;
}
if(i % 2 == 0)
num[i] = 7;
else
num[i+1] = 4;
k--;
}
}
for(int i = 1; i <= n; i++)
printf("%d", num[i]);
puts("");
return 0;
}