description: 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  
      
solution: #include <bits/stdc++.h>
  const double PI = 3.1415926535897932384626433;
  using namespace std;
  struct edge {
    long long to, cost;
    edge() {}
    edge(long long a, long long b) { to = a, cost = b; }
  };
  const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
  int mod = 1000000007;
  struct mint {
    int x;
    mint(long long y = 0) {
      if (y < 0 || y >= mod) y = (y % mod + mod) % mod;
      x = y;
    }
    mint(const mint& ope) { x = ope.x; }
    mint operator-() { return mint(-x); }
    mint operator+(const mint& ope) { return mint(x) += ope; }
    mint operator-(const mint& ope) { return mint(x) -= ope; }
    mint operator*(const mint& ope) { return mint(x) *= ope; }
    mint operator/(const mint& ope) { return mint(x) /= ope; }
    mint& operator+=(const mint& ope) {
      x += ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint& operator-=(const mint& ope) {
      x += mod - ope.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint& operator*=(const mint& ope) {
      long long tmp = x;
      tmp *= ope.x, tmp %= mod;
      x = tmp;
      return *this;
    }
    mint& operator/=(const mint& ope) {
      long long n = mod - 2;
      mint mul = ope;
      while (n) {
        if (n & 1) *this *= mul;
        mul *= mul;
        n >>= 1;
      }
      return *this;
    }
    mint inverse() { return mint(1) / *this; }
    bool operator==(const mint& ope) { return x == ope.x; }
    bool operator!=(const mint& ope) { return x != ope.x; }
    bool operator<(const mint& ope) { return x < ope.x; }
  };
  mint modpow(mint a, long long n) {
    if (n == 0) return mint(1);
    if (n % 2)
      return a * modpow(a, n - 1);
    else
      return modpow(a * a, n / 2);
  }
  istream& operator>>(istream& is, mint& ope) {
    long long t;
    is >> t, ope.x = t;
    return is;
  }
  ostream& operator<<(ostream& os, mint& ope) { return os << ope.x; }
  ostream& operator<<(ostream& os, const mint& ope) { return os << ope.x; }
  vector<mint> fact, fact_inv;
  void make_fact(int n) {
    fact.resize(n + 1), fact_inv.resize(n + 1);
    fact[0] = mint(1);
    for (long long i = (1); (i) <= (n); (i)++) fact[i] = fact[i - 1] * mint(i);
    fact_inv[n] = fact[n].inverse();
    for (long long i = (n - 1); (i) >= (0); (i)--)
      fact_inv[i] = fact_inv[i + 1] * mint(i + 1);
  }
  mint comb(int n, int k) {
    if (n < 0 || k < 0 || n < k) return mint(0);
    return fact[n] * fact_inv[k] * fact_inv[n - k];
  }
  mint perm(int n, int k) { return comb(n, k) * fact[k]; }
  vector<int> prime, pvec;
  void make_prime(int n) {
    prime.resize(n + 1);
    for (long long i = (2); (i) <= (n); (i)++) {
      if (prime[i]) continue;
      for (int j = i; j <= n; j += i) prime[j] = i;
    }
    for (long long i = (2); (i) <= (n); (i)++)
      if (prime[i] == i) pvec.push_back(i);
  }
  bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
  void mark() { cout << "*" << endl; }
  void yes() { cout << "YES" << endl; }
  void no() { cout << "NO" << endl; }
  long long floor(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return a / b;
    else
      return -((-a + b - 1) / b);
  }
  long long ceil(long long a, long long b) {
    if (b < 0) a *= -1, b *= -1;
    if (a >= 0)
      return (a + b - 1) / b;
    else
      return -((-a) / b);
  }
  long long modulo(long long a, long long b) {
    b = abs(b);
    return a - floor(a, b) * b;
  }
  long long sgn(long long x) {
    if (x > 0) return 1;
    if (x < 0) return -1;
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long digitnum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret++;
    return ret;
  }
  long long digitsum(long long x, long long b = 10) {
    long long ret = 0;
    for (; x; x /= b) ret += x % b;
    return ret;
  }
  string lltos(long long x) {
    string ret;
    for (; x; x /= 10) ret += x % 10 + '0';
    reverse(ret.begin(), ret.end());
    return ret;
  }
  long long stoll(string& s) {
    long long ret = 0;
    for (auto c : s) ret *= 10, ret += c - '0';
    return ret;
  }
  template <typename T>
  void uniq(T& vec) {
    sort(vec.begin(), vec.end());
    vec.erase(unique(vec.begin(), vec.end()), vec.end());
  }
  template <class S, class T>
  pair<S, T>& operator+=(pair<S, T>& s, const pair<S, T>& t) {
    s.first += t.first, s.second += t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T>& operator-=(pair<S, T>& s, const pair<S, T>& t) {
    s.first -= t.first, s.second -= t.second;
    return s;
  }
  template <class S, class T>
  pair<S, T> operator+(const pair<S, T>& s, const pair<S, T>& t) {
    return pair<S, T>(s.first + t.first, s.second + t.second);
  }
  template <class S, class T>
  pair<S, T> operator-(const pair<S, T>& s, const pair<S, T>& t) {
    return pair<S, T>(s.first - t.first, s.second - t.second);
  }
  template <typename T>
  ostream& operator<<(ostream& os, vector<T>& vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << (i + 1 == vec.size() ? "" : " ");
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& vec) {
    for (long long i = 0; (i) < (long long)(vec).size(); (i)++)
      os << vec[i] << (i + 1 == vec.size() ? "" : " ");
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, deque<T>& deq) {
    for (long long i = 0; (i) < (long long)(deq).size(); (i)++)
      os << deq[i] << (i + 1 == deq.size() ? "" : " ");
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, const pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& os, map<T, U>& map_var) {
    for (typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end();
         itr++) {
      os << "(" << itr->first << ", " << itr->second << ")";
      itr++;
      if (itr != map_var.end()) os << ",";
      itr--;
    }
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, set<T>& set_var) {
    for (typename set<T>::iterator itr = set_var.begin(); itr != set_var.end();
         itr++) {
      os << *itr;
      ++itr;
      if (itr != set_var.end()) os << " ";
      itr--;
    }
    return os;
  }
  template <typename T>
  ostream& operator<<(ostream& os, multiset<T>& set_var) {
    for (typename multiset<T>::iterator itr = set_var.begin();
         itr != set_var.end(); itr++) {
      os << *itr;
      ++itr;
      if (itr != set_var.end()) os << " ";
      itr--;
    }
    return os;
  }
  template <typename T>
  void outa(T a[], long long s, long long t) {
    for (long long i = (s); (i) <= (t); (i)++) {
      cout << a[i];
      if (i < t) cout << " ";
    }
    cout << endl;
  }
  void dump_func() { cout << endl; }
  template <class Head, class... Tail>
  void dump_func(Head&& head, Tail&&... tail) {
    cout << head;
    if (sizeof...(Tail) > 0) cout << " ";
    dump_func(std::move(tail)...);
  }
  long long h, w, Q;
  char c[1000005];
  long long sum[1000005];
  int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> h >> w;
    for (long long y = (0); (y) <= (h - 1); (y)++)
      for (long long x = (0); (x) <= (w - 1); (x)++) cin >> c[y * w + x];
    for (long long y = (0); (y) <= (h - 1); (y)++)
      for (long long x = (0); (x) <= (w - 1); (x)++) {
        if (y - 1 >= 0 && x + 1 < w && c[y * w + x] == 'X' &&
            c[(y - 1) * w + (x + 1)] == 'X')
          sum[x + 1]++;
      }
    for (long long i = (1); (i) <= (w); (i)++) sum[i] += sum[i - 1];
    cin >> Q;
    long long l, r;
    for (long long i = (1); (i) <= (Q); (i)++) {
      cin >> l >> r;
      if (sum[r - 1] - sum[l - 1] == 0)
        cout << "YES"
             << "\n";
      else
        cout << "NO"
             << "\n";
    }
    return 0;
  }
  