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/**
 * Calculate the number of ways to tile a 2×n board with dominos and trominos.
 * Dynamic Programming with Multiple States approach.
 *
 * @param {number} n - Width of the board
 * @return {number} - Number of ways to tile the board
 */
function numTilings(n) {
  const MOD = 1000000007;
 
  // Base cases
  if (n === 1) {
    return 1;
  }
 
  // Initialize DP arrays
  const dp = new Array(n + 1).fill(0);   // dp[i] = ways to fully tile a 2×i board
  const dp2 = new Array(n + 1).fill(0);  // dp2[i] = ways to tile a 2×i board with one protruding square
 
  // Base cases
  dp[0] = 1;  // Empty board has one way to tile (do nothing)
  dp[1] = 1;  // 2×1 board has one way to tile (vertical domino)
  dp2[0] = 0; // Empty board has no way to tile with a protruding square
  dp2[1] = 1; // 2×1 board has one way to tile with a protruding square
 
  // Fill DP arrays
  for (let i = 2; i <= n; i++) {
    // dp2[i] = ways to tile with a protruding square
    // = ways to place a horizontal domino on a board with a protruding square
    // + ways to place a tromino on a fully tiled board
    dp2[i] = (dp2[i - 1] + dp[i - 2]) % MOD;
 
    // dp[i] = ways to fully tile a 2×i board
    // = ways to place a vertical domino on a fully tiled board
    // + ways to place two horizontal dominos on a fully tiled board
    // + ways to place a tromino on a board with a protruding square (2 orientations)
    dp[i] = (dp[i - 1] + dp[i - 2] + 2 * dp2[i - 1]) % MOD;
  }
 
  return dp[n];
}
 
/**
 * Calculate the number of ways to tile a 2×n board with dominos and trominos.
 * Dynamic Programming with Recurrence Relation approach.
 *
 * @param {number} n - Width of the board
 * @return {number} - Number of ways to tile the board
 */
function numTilingsRecurrence(n) {
  const MOD = 1000000007;
 
  // Base cases
  if (n === 1) {
    return 1;
  }
  if (n === 2) {
    return 2;
  }
  if (n === 3) {
    return 5;
  }
 
  // Initialize DP array
  const dp = new Array(n + 1).fill(0);
 
  // Base cases
  dp[1] = 1;
  dp[2] = 2;
  dp[3] = 5;
 
  // Fill DP array using the recurrence relation
  for (let i = 4; i <= n; i++) {
    dp[i] = (2 * dp[i - 1] % MOD + dp[i - 3]) % MOD;
  }
 
  return dp[n];
}
 
/**
 * Calculate the number of ways to tile a 2×n board with dominos and trominos.
 * Matrix Exponentiation approach.
 *
 * @param {number} n - Width of the board
 * @return {number} - Number of ways to tile the board
 */
function numTilingsMatrix(n) {
  const MOD = 1000000007;
 
  // Base cases
  if (n === 1) {
    return 1;
  }
  if (n === 2) {
    return 2;
  }
  if (n === 3) {
    return 5;
  }
 
  // Define the base matrix
  const base = [
    [2, 0, 1],
    [1, 0, 0],
    [0, 1, 0]
  ];
 
  // Compute base^(n-3)
  const result = matrixPower(base, n - 3, MOD);
 
  // Multiply with the initial values [5, 2, 1]
  return (result[0][0] * 5 + result[0][1] * 2 + result[0][2] * 1) % MOD;
}
 
/**
 * Compute the power of a matrix using binary exponentiation.
 *
 * @param {number[][]} matrix - The base matrix
 * @param {number} n - The exponent
 * @param {number} mod - The modulo value
 * @return {number[][]} - The resulting matrix
 */
function matrixPower(matrix, n, mod) {
  // Base case: matrix^0 = identity matrix
  if (n === 0) {
    const size = matrix.length;
    const result = Array(size).fill().map(() => Array(size).fill(0));
    for (let i = 0; i < size; i++) {
      result[i][i] = 1;
    }
    return result;
  }
 
  // If n is odd, compute matrix^(n-1) * matrix
  if (n % 2 === 1) {
    const result = matrixPower(matrix, n - 1, mod);
    return multiplyMatrices(result, matrix, mod);
  }
 
  // If n is even, compute (matrix^(n/2))^2
  const half = matrixPower(matrix, Math.floor(n / 2), mod);
  return multiplyMatrices(half, half, mod);
}
 
/**
 * Multiply two matrices.
 *
 * @param {number[][]} a - First matrix
 * @param {number[][]} b - Second matrix
 * @param {number} mod - The modulo value
 * @return {number[][]} - The product of the two matrices
 */
function multiplyMatrices(a, b, mod) {
  const rowsA = a.length;
  const colsA = a[0].length;
  const colsB = b[0].length;
 
  const result = Array(rowsA).fill().map(() => Array(colsB).fill(0));
 
  for (let i = 0; i < rowsA; i++) {
    for (let j = 0; j < colsB; j++) {
      for (let k = 0; k < colsA; k++) {
        result[i][j] = (result[i][j] + a[i][k] * b[k][j]) % mod;
      }
    }
  }
 
  return result;
}
 
// Test cases
console.log(numTilings(1));  // 1
console.log(numTilings(2));  // 2
console.log(numTilings(3));  // 5
console.log(numTilings(4));  // 11
console.log(numTilings(5));  // 24
 
console.log(numTilingsRecurrence(1));  // 1
console.log(numTilingsRecurrence(2));  // 2
console.log(numTilingsRecurrence(3));  // 5
console.log(numTilingsRecurrence(4));  // 11
console.log(numTilingsRecurrence(5));  // 24
 
console.log(numTilingsMatrix(1));  // 1
console.log(numTilingsMatrix(2));  // 2
console.log(numTilingsMatrix(3));  // 5
console.log(numTilingsMatrix(4));  // 11
console.log(numTilingsMatrix(5));  // 24