CodeForces - 1183E Subsequences (easy version) （字符串bfs）

The only difference between the easy and the hard versions is constraints.

A subsequence is a string that can be derived from another string by deleting some or no symbols without changing the order of the remaining symbols. Characters to be deleted are not required to go successively, there can be any gaps between them. For example, for the string "abaca" the following strings are subsequences: "abaca", "aba", "aaa", "a" and "" (empty string). But the following strings are not subsequences: "aabaca", "cb" and "bcaa".

You are given a string ss consisting of nn lowercase Latin letters.

In one move you can take any subsequence tt of the given string and add it to the set SS. The set SS can't contain duplicates. This move costs n−|t|n−|t|, where |t||t| is the length of the added subsequence (i.e. the price equals to the number of the deleted characters).

Your task is to find out the minimum possible total cost to obtain a set SS of size kk or report that it is impossible to do so.

Input

The first line of the input contains two integers nn and kk (1≤n,k≤1001≤n,k≤100) — the length of the string and the size of the set, correspondingly.

The second line of the input contains a string ss consisting of nn lowercase Latin letters.

Output

Print one integer — if it is impossible to obtain the set SS of size kk, print -1. Otherwise, print the minimum possible total cost to do it.

Examples

Input

4 5
asdf

Output

4

Input

5 6
aaaaa

Output

15

Input

5 7
aaaaa

Output

-1

Input

10 100
ajihiushda

Output

233

Note

In the first example we can generate SS = { "asdf", "asd", "adf", "asf", "sdf" }. The cost of the first element in SS is 00 and the cost of the others is 11. So the total cost of SS is 44.

 

 

題意：

給你一個長度為n的字符串，找出其中k個不同子序列（可不連續），使得代價（刪除字符數）最小。

思路：

如果通過dfs找遍所有子串將會有2^100種可能，顯然行不通。

可以將字符串抽象成圖，字符是一個個節點。利用bfs與set結合，隊列存儲當前字符串，每次刪除一個字符，若set不存在，更新隊列與set。

bfs層先能夠從刪除少的字符串開始，保證了代價最小效率最優。

 

官方題解

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
#endif
    
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    
    int ans = 0;
    queue<string> q;
    set<string> st;
    q.push(s);
    st.insert(s);
    while (!q.empty() && int(st.size()) < k) {
        string v = q.front();
        q.pop();
        for (int i = 0; i < int(v.size()); ++i) {
            string nv = v;
            nv.erase(i, 1);
            if (!st.count(nv) && int(st.size()) + 1 <= k) {
                q.push(nv);
                st.insert(nv);
                ans += n - nv.size();
            }
        }
    }
    
    if (int(st.size()) < k) cout << -1 << endl;
    else cout << ans << endl;
    
    return 0;
}