Problem: Apple adventure

square1001 and E869120 got lost in the grid world of $ H $ rows and $ W $ rows!

Said the god of this world.

"When you collect $ K $ apples and they meet, you'll be back in the original world."

Upon hearing this word, square1001 decided to collect more than $ K $ of apples and head to the square where E869120 is.




Here, each cell in the grid is represented as follows.

's': square1001 This is the square where you are.

'e': E869120 This is the square where you are.

'a': A square with one apple on it. You can get an apple the first time you visit this trout. There are no more than 20 squares on this grid.

'#': It's a wall. You cannot visit this square.

'.': A square with nothing. You can visit this square.




square1001 You try to achieve your goal by repeatedly moving from the square you are in to the squares that are adjacent to each other up, down, left, and right. However, you cannot get out of the grid.

square1001 Find the minimum number of moves you need to achieve your goal.

However, it is assumed that E869120 will not move. Also, square1001 shall be capable of carrying $ K $ or more of apples.

If the goal cannot be achieved, output "-1".

input

Input is given from standard input in the following format.

Let $ A_ {i, j} $ be the characters in the $ i $ square from the top of the grid and the $ j $ square from the left.


$ H $ $ W $ $ K $
$ A_ {1,1} A_ {1,2} A_ {1,3} \ cdots A_ {1, W} $
$ A_ {2,1} A_ {2,2} A_ {2,3} \ cdots A_ {2, W} $
$ A_ {3,1} A_ {3,2} A_ {3,3} \ cdots A_ {3, W} $
$ \ ldots $
$ A_ {H, 1} A_ {H, 2} A_ {H, 3} \ cdots A_ {H, W} $


output

square1001 Find the minimum number of moves you need to reach your goal. However, if this is not possible, output "-1".

However, insert a line break at the end.

Constraint

* $ 1 \ leq H \ leq 1000 $
* $ 1 \ leq W \ leq 1000 $
* $ 1 \ leq K \ leq 20 $
* $ H, W, K $ are integers.
* $ A_ {i, j} $ is one of's','e','a','#','.'.
* The grid contains only one's' and one'e'.
* The number of'a'in the grid is greater than or equal to $ K $ and less than or equal to $ 20 $.



Input example 1


5 5 2
s .. # a
. # ...
a # e. #
... # a
. # ...


Output example 1


14


Input example 2


7 7 3
.......
.s ... a.
a ## ... a
.. ### ..
.a # e # .a
. ### ..
a .. # .. a


Output example 2


-1


If the purpose cannot be achieved, output "-1".

Input example 3


12 12 10
. ##### ......
.## ..... # ...
.... a.a # a .. #
. # .. # a ......
..... a # s ..
..a ###. ##. #
.e #. #. #. # A ..
.. # a # ..... #.
.. ## a ......
.a ... a.a .. #.
a .... # a.aa ..
... a. # ... # a.


Output example 3


30






Example

Input

5 5 2
s..#a
.#...
a#e.#
...#a
.#...


Output

14

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
const lint mod = 998244353;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl((lint)(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
lint power(lint base, lint exponent, lint module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ lint root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr first, itr last){ for(auto i = first; i != last; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<lint> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } lint C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };
template<class itr> bool next_sequence(itr first, itr last, int max_bound){ itr now = last; while(now != first){ now--; (*now)++; if((*now) == max_bound){ (*now) = 0; }else{ return true; } } return false; }
template<class itr, class itr2> bool next_sequence2(itr first, itr last, itr2 first2, itr2 last2){ itr now = last; itr2 now2 = last2; while(now != first){ now--, now2--; (*now)++; if((*now) == (*now2)){ (*now) = 0; }else{ return true; } } return false; }

string grid[1003];
int dist[1003][1003];

int H, W, K;

int apple_size;
vector<vector<int>> apple;

int dp[21][1 << 21];

int move(int now, int where){
    if(dp[now][where] != -1){
        return dp[now][where];
    }
    if(bitcount(where) == K){
        if(apple[now][apple_size + 1] != -1){
            return dp[now][where] = apple[now][apple_size + 1];
        }else{
            return dp[now][where] = 1e9;
        }
    }
    int ans = 1e9;
    for(int i = 0; i < apple_size; i++){
        if(apple[now][i] == -1){
            continue;
        }
        if((where >> i) & 1){
            continue;
        }
        ans = min(ans, move(i, where | (1 << i)) + apple[now][i]);
    }
    return dp[now][where] = ans;
}

int main(){
    scanf("%d %d %d\n", &H, &W, &K);
    for(int i = 0; i < H; i++){
        grid[i].resize(W);
        for(int j = 0; j < W; j++){
            scanf("%c", &grid[i][j]);
        }
        cin.ignore();
    }
    vector<position> pos;
    position start, goal;
    for(int i = 0; i < H; i++){
        for(int j = 0; j < W; j++){
            if(grid[i][j] == 'a'){
                pos.push_back({i, j});
            }else if(grid[i][j] == 's'){
                start = {i, j};
            }else if(grid[i][j] == 'e'){
                goal = {i, j};
            }
        }
    }
    apple_size = int(pos.size());
    pos.push_back(start);
    pos.push_back(goal);
    apple.resize(pos.size(), vector<int>(pos.size()));
    int i_cnt = 0;
    for(position i: pos){
        for(int i = 0; i < 1003; i++){
            for(int j = 0; j < 1003; j++){
                dist[i][j] = -1;
            }
        }
        queue<position> pathes;
        pathes.push(i);
        dist[i.y][i.x] = 0;
        while(!pathes.empty()){
            auto top = pathes.front();
            pathes.pop();
            for(position j: mv){
                int next_y = top.y + j.y;
                int next_x = top.x + j.x;
                if(0 <= next_y && next_y < H && 0 <= next_x && next_x < W){
                    if(dist[next_y][next_x] == -1 && grid[next_y][next_x] != '#'){
                        dist[next_y][next_x] = dist[top.y][top.x] + 1;
                        pathes.push({next_y, next_x});
                    }
                }
            }
        }
        int j_cnt = 0;
        for(position j: pos){
            apple[i_cnt][j_cnt] = dist[j.y][j.x];
            j_cnt++;
        }
        i_cnt++;
    }
    /*for(auto i: apple){
        array_output(all(i));
    }*/
    memset(dp, -1, sizeof(dp));
    int ans = move(apple_size, 0);
    if(ans > 1e9 - 100){
        printf("%d\n", -1);
    }else{
        printf("%d\n", ans);
    }
    /*cout << ans << endl;
    cout << apple_size << " " << apple.size() << endl;*/
}