#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const long double pi = 2 * acos(0.0);
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<string> a(n);
  for (auto &element : a) cin >> element;
  int q;
  vector<vector<int> > ans(n, vector<int>(m));
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      if (a[i - 1][j] == 'X' && a[i][j - 1] == 'X') ans[i][j] = 1;
      ans[i][j] += ans[i - 1][j] + ans[i][j - 1] - ans[i - 1][j - 1];
    }
  }
  cin >> q;
  while (q--) {
    int x1, x2;
    cin >> x1 >> x2;
    x1--;
    x2--;
    if (x2 - x1 >= 1) {
      int curr = ans[n - 1][x2] - ans[n - 1][x1];
      if (curr > 0) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
      }
    } else {
      cout << "YES\n";
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

The problem statement looms below, filling you with determination.

Consider a grid in which some cells are empty and some cells are filled. Call a cell in this grid exitable if, starting at that cell, you can exit the grid by moving up and left through only empty cells. This includes the cell itself, so all filled in cells are not exitable. Note that you can exit the grid from any leftmost empty cell (cell in the first column) by going left, and from any topmost empty cell (cell in the first row) by going up.

Let's call a grid determinable if, given only which cells are exitable, we can exactly determine which cells are filled in and which aren't.

You are given a grid a of dimensions n × m , i. e. a grid with n rows and m columns. You need to answer q queries (1 ≤ q ≤ 2 ⋅ 10^5). Each query gives two integers x_1, x_2 (1 ≤ x_1 ≤ x_2 ≤ m) and asks whether the subgrid of a consisting of the columns x_1, x_1 + 1, …, x_2 - 1, x_2 is determinable.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 10^6, nm ≤ 10^6) — the dimensions of the grid a.

n lines follow. The y-th line contains m characters, the x-th of which is 'X' if the cell on the intersection of the the y-th row and x-th column is filled and "." if it is empty.

The next line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

q lines follow. Each line contains two integers x_1 and x_2 (1 ≤ x_1 ≤ x_2 ≤ m), representing a query asking whether the subgrid of a containing the columns x_1, x_1 + 1, …, x_2 - 1, x_2 is determinable.

Output

For each query, output one line containing "YES" if the subgrid specified by the query is determinable and "NO" otherwise. The output is case insensitive (so "yEs" and "No" will also be accepted).

Example

Input


4 5
..XXX
...X.
...X.
...X.
5
1 3
3 3
4 5
5 5
1 5


Output


YES
YES
NO
YES
NO

Note

For each query of the example, the corresponding subgrid is displayed twice below: first in its input format, then with each cell marked as "E" if it is exitable and "N" otherwise.

For the first query: 
    
    
      
    ..X EEN  
    ... EEE  
    ... EEE  
    ... EEE  
    
    
    
      
      
    

For the second query: 
    
    
      
    X N  
    . E  
    . E  
    . E  
    

Note that you can exit the grid by going left from any leftmost cell (or up from any topmost cell); you do not need to reach the top left corner cell to exit the grid.
    
    
      
      
    

For the third query: 
    
    
      
    XX NN  
    X. NN  
    X. NN  
    X. NN  
    

This subgrid cannot be determined only from whether each cell is exitable, because the below grid produces the above "exitability grid" as well: 
    
    
      
    XX  
    XX  
    XX  
    XX  
    
    
    
      
      
    

For the fourth query: 
    
    
      
    X N  
    . E  
    . E  
    . E  
    
    
    
      
      
    

For the fifth query: 
    
    
      
    ..XXX EENNN  
    ...X. EEENN  
    ...X. EEENN  
    ...X. EEENN  
    

This query is simply the entire grid. It cannot be determined only from whether each cell is exitable because the below grid produces the above "exitability grid" as well: 
    
    
      
    ..XXX  
    ...XX  
    ...XX  
    ...XX  
    

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const long double pi = 2 * acos(0.0);
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<string> a(n);
  for (auto &element : a) cin >> element;
  int q;
  vector<vector<int> > ans(n, vector<int>(m));
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      if (a[i - 1][j] == 'X' && a[i][j - 1] == 'X') ans[i][j] = 1;
      ans[i][j] += ans[i - 1][j] + ans[i][j - 1] - ans[i - 1][j - 1];
    }
  }
  cin >> q;
  while (q--) {
    int x1, x2;
    cin >> x1 >> x2;
    x1--;
    x2--;
    if (x2 - x1 >= 1) {
      int curr = ans[n - 1][x2] - ans[n - 1][x1];
      if (curr > 0) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
      }
    } else {
      cout << "YES\n";
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```