text: # Time:  O(logn)
  # Space: O(logn)
  
  # We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape.
  # These shapes may be rotated.
  #
  # XX  <- domino
  #
  # XX  <- "L" tromino
  # X
  # Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.
  #
  # (In a tiling, every square must be covered by a tile.
  # Two tilings are different if and only if there are two 4-directionally adjacent cells on the board
  # such that exactly one of the tilings has both squares occupied by a tile.)
  #
  # Example:
  # Input: 3
  # Output: 5
  # Explanation:
  # The five different ways are listed below, different letters indicates different tiles:
  # XYZ XXZ XYY XXY XYY
  # XYZ YYZ XZZ XYY XXY
  #
  # Note:
  # - N  will be in range [1, 1000].
  
  import itertools
  
  
  class Solution(object):
      def numTilings(self, N):
          """
          :type N: int
          :rtype: int
          """
          M = int(1e9+7)
  
          def matrix_expo(A, K):
              if K == 0:
                  return [[int(i==j) for j in xrange(len(A))] \
                          for i in xrange(len(A))]
              if K == 1:
                  return A
              if K % 2:
                  return matrix_mult(matrix_expo(A, K-1), A)
              B = matrix_expo(A, K//2)
              return matrix_mult(B, B)
  
          def matrix_mult(A, B):
              ZB = zip(*B)
              return [[sum(a*b for a, b in itertools.izip(row, col)) % M \
                       for col in ZB] for row in A]
  
          T = [[1, 0, 0, 1],  # #(|) = #(|) + #(=)
               [1, 0, 1, 0],  # #(「) = #(|) + #(L)
               [1, 1, 0, 0],  # #(L) = #(|) + #(「)
               [1, 1, 1, 0]]  # #(=) = #(|) + #(「) + #(L)
  
          return matrix_expo(T, N)[0][0]  # T^N * [1, 0, 0, 0]
  
  
  # Time:  O(n)
  # Space: O(1)
  class Solution2(object):
      def numTilings(self, N):
          """
          :type N: int
          :rtype: int
          """
          # Prove:
          # dp[n] = dp[n-1](|) + dp[n-2](=) + 2*(dp[n-3](「」) + ... + d[0](「 = ... = 」))
          #       = dp[n-1] + dp[n-2] + dp[n-3] + dp[n-3] + 2*(dp[n-4] + ... + d[0])
          #       = dp[n-1] + dp[n-3] + (dp[n-2] + dp[n-3] + 2*(dp[n-4] + ... + d[0])
          #       = dp[n-1] + dp[n-3] + dp[n-1]
          #       = 2*dp[n-1] + dp[n-3]
          M = int(1e9+7)
          dp = [1, 1, 2]
          for i in xrange(3, N+1):
              dp[i%3] = (2*dp[(i-1)%3]%M + dp[(i-3)%3])%M
          return dp[N%3]
  