You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). A nonnegative integer A_{i,j} is written for each square (i,j).

You choose some of the squares so that each row and column contains at most K chosen squares. Under this constraint, calculate the maximum value of the sum of the integers written on the chosen squares. Additionally, calculate a way to choose squares that acheives the maximum.

Constraints

* 1 \leq N \leq 50
* 1 \leq K \leq N
* 0 \leq A_{i,j} \leq 10^9
* All values in Input are integer.

Input

Input is given from Standard Input in the following format:


N K
A_{1,1} A_{1,2} \cdots A_{1,N}
A_{2,1} A_{2,2} \cdots A_{2,N}
\vdots
A_{N,1} A_{N,2} \cdots A_{N,N}


Output

On the first line, print the maximum value of the sum of the integers written on the chosen squares.

On the next N lines, print a way that achieves the maximum.

Precisely, output the strings t_1,t_2,\cdots,t_N, that satisfies t_{i,j}=`X` if you choose (i,j) and t_{i,j}=`.` otherwise.

You may print any way to choose squares that maximizes the sum.

Examples

Input

3 1
5 3 2
1 4 8
7 6 9


Output

19
X..
..X
.X.


Input

3 2
10 10 1
10 10 1
1 1 10


Output

50
XX.
XX.
..X
from heapq import heappop, heappush


class MinCostFlow:
    def __init__(self, n):
        self.n = n
        self.graph = [[] for _ in range(n)]
        self.pos = []
        self.pv = []
        self.pe = []
        self.dual = []

    def add_edge(self, fr, to, cap, cost):
        m = len(self.pos)
        self.pos.append((fr, len(self.graph[fr])))
        self.graph[fr].append([to, len(self.graph[to]), cap, cost])
        self.graph[to].append([fr, len(self.graph[fr]) - 1, 0, -cost])
        return m

    def get_edge(self, idx):
        to, rev, cap, cost = self.graph[self.pos[idx][0]][self.pos[idx][1]]
        rev_to, rev_rev, rev_cap, rev_cost = self.graph[to][rev]
        return self.pos[idx][0], to, cap + rev_cap, rev_cap, cost

    def edges(self):
        for i in range(len(self.pos)):
            yield self.get_edge(i)

    def dual_ref(self, s, t):
        dist = [2 ** 63 - 1] * self.n
        dist[s] = 0
        vis = [0] * self.n
        self.pv = [-1] * self.n
        self.pe = [-1] * self.n
        queue = []
        heappush(queue, (0, s))
        while queue:
            k, v = heappop(queue)
            if vis[v]:
                continue
            vis[v] = True
            if v == t:
                break
            for i in range(len(self.graph[v])):
                to, rev, cap, cost = self.graph[v][i]
                if vis[to] or cap == 0:
                    continue
                cost += self.dual[v] - self.dual[to]
                if dist[to] - dist[v] > cost:
                    dist[to] = dist[v] + cost
                    self.pv[to] = v
                    self.pe[to] = i
                    heappush(queue, (dist[to], to))
        if not vis[t]:
            return False
        for v in range(self.n):
            if not vis[v]:
                continue
            self.dual[v] -= dist[t] - dist[v]
        return True

    def flow(self, s, t):
        return self.flow_with_limit(s, t, 2 ** 63 - 1)

    def flow_with_limit(self, s, t, limit):
        return self.slope_with_limit(s, t, limit)[-1]

    def slope(self, s, t):
        return self.slope_with_limit(s, t, 2 ** 63 - 1)

    def slope_with_limit(self, s, t, limit):
        flow = 0
        cost = 0
        prev_cost = -1
        res = [(flow, cost)]
        self.dual = [0] * self.n
        while flow < limit:
            if not self.dual_ref(s, t):
                break
            c = limit - flow
            v = t
            while v != s:
                c = min(c, self.graph[self.pv[v]][self.pe[v]][2])
                v = self.pv[v]
            v = t
            while v != s:
                to, rev, cap, _ = self.graph[self.pv[v]][self.pe[v]]
                self.graph[self.pv[v]][self.pe[v]][2] -= c
                self.graph[v][rev][2] += c
                v = self.pv[v]
            d = -self.dual[s]
            flow += c
            cost += c * d
            if prev_cost == d:
                res.pop()
            res.append((flow, cost))
            prev_cost = cost
        return res


N, K = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(N)]

mcf = MinCostFlow(2 * N + 2)
start = 2 * N
end = 2 * N + 1

MAX = 10 ** 9

mcf.add_edge(start, end, N * K, MAX)

for i in range(N):
    mcf.add_edge(start, i, K, 0)
    mcf.add_edge(N + i, end, K, 0)

    for j in range(N):
        mcf.add_edge(i, N + j, 1, MAX - A[i][j])

print(N * K * MAX - mcf.flow_with_limit(start, end, N * K)[1])

for fr, to, cap, flow, cost in mcf.edges():
    if fr == start or to == end:
        continue
    A[fr][to-N] = 'X' if flow == 1 else '.'

for r in A:
    print(''.join(r))
