<p>You are given an <code>m x n</code> integer matrix <code>grid</code>, where you can move from a cell to any adjacent cell in all <code>4</code> directions.</p>

<p>Return <em>the number of <strong>strictly</strong> <strong>increasing</strong> paths in the grid such that you can start from <strong>any</strong> cell and end at <strong>any</strong> cell. </em>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two paths are considered different if they do not have exactly the same sequence of visited cells.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/10/griddrawio-4.png" style="width: 181px; height: 121px;" />
<pre>
<strong>Input:</strong> grid = [[1,1],[3,4]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -&gt; 3], [1 -&gt; 4], [3 -&gt; 4].
- Paths with length 3: [1 -&gt; 3 -&gt; 4].
The total number of paths is 4 + 3 + 1 = 8.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1],[2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [2].
- Paths with length 2: [1 -&gt; 2].
The total number of paths is 2 + 1 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

template <class T>
ostream &operator<<(ostream &os, const vector<vector<T>> &mat) {
  for (auto &line : mat) {
    os << "[";
    for (auto &item : line) {
      os << item << ",";
    }
    os << "]\n";
  }
  return os;
}

class Solution {
public:
  int countPaths(vector<vector<int>> &matrix) {
    int n = matrix.size();
    int m = matrix.front().size();
    vector<vector<int>> memo(n, vector<int>(m, -1));
    vector<vector<bool>> visited(n, vector<bool>(m, false));
    int result = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        result += dp(matrix, memo, visited, n, m, i, j);
        result %= 1000000007L;
      }
    }
    //cout << memo << endl;
    return result;
  }

  bool inBound(int i, int j, int n, int m) {
    return i >= 0 && i < n && j >= 0 && j < m;
  }

  int dp(const vector<vector<int>> &matrix, vector<vector<int>> &memo,
         vector<vector<bool>> &visited, int n, int m, int i, int j) {
    if (!inBound(i, j, n, m)) {
      return 0;
    }
    if (memo[i][j] != -1) {
      return memo[i][j];
    }
    assert(!visited[i][j]);
    visited[i][j] = true;
    vector<pair<int, int>> moves = {make_pair(-1, 0), make_pair(1, 0),
                                    make_pair(0, -1), make_pair(0, 1)};
    int result = 1;
    for (auto &move : moves) {
      int dx = move.first, dy = move.second;
      if (inBound(i + dy, j + dx, n, m) &&
          matrix[i + dy][j + dx] > matrix[i][j]) {
        result += dp(matrix, memo, visited, n, m, i + dy, j + dx);
        result %= 1000000007L;
      }
    }
    visited[i][j] = false;
    return memo[i][j] = result;
  }
};

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