output_description: On the first line, print the maximum number of tiles that can be placed.
  
  On the next N lines, print a configulation that achieves the maximum.
  Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following
  way.
  
    * t_i is initialized to S_i.
    * For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
    * For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
  
  See samples for further information.
  
  You may print any configulation that maximizes the number of tiles.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
      S_1
      S_2
      \vdots
      S_N
attempt: class mf_graph:
      def __init__(self, n=0):
          self._n = n
          self.g = [[] for _ in range(n)]
          self.pos = []
  
      def add_edge(self, frm, to, cap):
          m = len(self.pos)
          e1 = mf_graph._edge(to, cap)
          e2 = mf_graph._edge(frm, 0)
          e1.rev = e2
          e2.rev = e1
          self.pos.append(e1)
          self.g[frm].append(e1)
          self.g[to].append(e2)
          return m
  
      class edge:
          def __init__(self, frm, to, cap, flow):
              self.frm = frm
              self.to = to
              self.cap = cap
              self.flow = flow
  
          def __iter__(self):
              yield self.frm
              yield self.to
              yield self.cap
              yield self.flow
  
      def get_edge(self, i):
          e1 = self.pos[i]
          e2 = e1.rev
          return mf_graph.edge(e2.to, e1.to, e1.cap + e2.cap, e2.cap)
  
      def edges(self):
          return [self.get_edge(i) for i in range(len(self.pos))]
  
      def change_edge(self, i, new_cap, new_flow):
          e = self.pos[i]
          e.cap = new_cap - new_flow
          e.rev.cap = new_flow
  
      def flow(self, s, t, flow_limit=0xFFFFFFFFFFFFFFF):
          g = self.g
          flow = 0
          es = [[] for _ in range(self._n)]
          while flow < flow_limit:
              level = [-1] * self._n
              level[s] = 0
              que = [None] * self._n
              ql = 0
              qr = 1
              que[0] = s
              unreached = s != t
              lvl = 1
              for p in es:
                  p.clear()
              while unreached and ql < qr:
                  nqr = qr
                  while ql < qr:
                      v = que[ql]
                      ql += 1
                      for e in g[v]:
                          to = e.to
                          if e.cap and not 0 <= level[to] < lvl:
                              es[to].append(e.rev)
                              if level[to] < 0:
                                  level[to] = lvl
                                  if to == t:
                                      unreached = False
                                  que[nqr] = to
                                  nqr += 1
                  qr = nqr
                  lvl += 1
              if unreached:
                  return flow
              stack = []
              v = t
              up = flow_limit - flow
              res = 0
              while True:
                  if v == s or not es[v]:
                      if v == s:
                          res = up
                      while stack:
                          tmp = res
                          e, up, res = stack.pop()
                          e.cap -= tmp
                          e.rev.cap += tmp
                          res += tmp
                          if res < up:
                              v = e.to
                              break
                      else:
                          flow += res
                          break
                  else:
                      e = es[v].pop()
                      stack.append((e.rev, up, res))
                      v = e.to
                      up = min(up, e.rev.cap)
                      res = 0
          return flow
  
      def min_cut(self, s):
          visited = [False] * self._n
          que = [None] * self._n
          ql = 0
          qr = 1
          que[0] = s
          visited[s] = True
          while ql < qr:
              p = que[ql]
              ql += 1
              for e in self.g[p]:
                  if e.cap and not visited[e.to]:
                      visited[e.to] = True
                      que[qr] = e.to
                      qr += 1
          return visited
  
      class _edge:
          def __init__(self, to, cap):
              self.to = to
              self.cap = cap
  
  
  n, m = map(int, input().split())
  s = [input() for _ in range(n)]
  g = mf_graph(n * m + 2)
  for i in range(n):
      for j in range(i % 2, m, 2):
          if s[i][j] == "#":
              continue
          for di, dj in (1, 0), (0, 1), (-1, 0), (0, -1):
              x, y = i + di, j + dj
              if 0 <= x < n and 0 <= y < m and s[x][y] == ".":
                  g.add_edge(i * m + j, x * m + y, 1)
  for i in range(n):
      for j in range(m):
          if s[i][j] == "#":
              continue
          if (i + j) % 2 == 0:
              g.add_edge(n * m, i * m + j, 1)
          else:
              g.add_edge(i * m + j, n * m + 1, 1)
  print(g.flow(n * m, n * m + 1))
  ans = [list(si) for si in s]
  for frm, to, cap, flow in g.edges():
      if flow and frm < n * m and to < n * m:
          i, j = frm // m, frm % m
          x, y = to // m, to % m
          if j == y:
              if i > x:
                  i, j, x, y = x, y, i, j
              ans[i][j], ans[x][y] = "v^"
          else:
              if j > y:
                  i, j, x, y = x, y, i, j
              ans[i][j], ans[x][y] = "><"
  
  for a in ans:
      print("".join(a))
  
problem_description: Statement
  
  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). Some of the squares contain an object.
  All the remaining squares are empty. The state of the grid is represented by
  strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}=
  `#` and is empty if S_{i,j}= `.`.
  
  Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically
  or horizontally to cover two adjacent empty squares. Tiles must not stick out
  of the grid, and no two different tiles may intersect. Tiles cannot occupy the
  square with an object.
  
  Calculate the maximum number of tiles that can be placed and any configulation
  that acheives the maximum.
samples: [{"input": "3 3\n    #..\n    ..#\n    ...", "output": "3\n    #><\n    vv#\n    ^^.\n    \n\nThe following output is also treated as a correct answer.\n\n    \n    \n    3\n    #><\n    v.#\n    ^><"}]