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  
    
import java.io.*;
import java.util.*;

public class C {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static FastReader s = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    private static int[] rai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        return arr;
    }

    private static int[][] rai(int n, int m) {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextInt();
            }
        }
        return arr;
    }

    private static long[] ral(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextLong();
        }
        return arr;
    }

    private static long[][] ral(int n, int m) {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextLong();
            }
        }
        return arr;
    }

    private static int ri() {
        return s.nextInt();
    }

    private static long rl() {
        return s.nextLong();
    }

    private static String rs() {
        return s.next();
    }

    static long gcd(long a,long b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static int gcd(int a,int b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static boolean isPrime(int n) {
        //check if n is a multiple of 2
        if (n % 2 == 0) return false;
        //if not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long mpow(long base,long pow)
    {
        long res=1;
        while(pow>0)
        {
            if(pow%2==1)
            {
                res=(res*base)%MOD;
            }
            pow>>=1;
            base=(base*base)%MOD;
        }
        return res;
    }

    static TreeSet<Long> set= new TreeSet<>();
    static void sieveOfEratosthenes(int n)
    {
        // Create a boolean array
        // "prime[0..n]" and
        // initialize all entries
        // it as true. A value in
        // prime[i] will finally be
        // false if i is Not a
        // prime, else true.
        boolean[] prime = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++)
        {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p])
            {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        // Print all prime numbers
        for (int i = 2; i <= n; i++)
        {
            if (prime[i])
            {
                for(int pow = 1;pow<=10;pow++) {
                    long val = (long) Math.pow((long)i, pow);
                    set.add(val);
                    if(val>1000)
                    {
                        break;
                    }
                }
            }
        }
    }

    static long smallestDivisor(long n)
    {
        // if divisible by 2
        if (n % 2 == 0)
            return 2;

        // iterate from 3 to sqrt(n)
        for (long i = 3; i * i <= n; i += 2) {
            if (n % i == 0)
                return i;
        }

        return n;
    }


    static long power(long x, long y, int p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r)
    {

        if (n<r)
            return 0;
        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r


        return (fac[n] * modInverse(fac[r], MOD)
                % MOD * modInverse(fac[n - r], MOD)
                % MOD)
                % MOD;
    }
    static int MOD = (int)(1e9+7);
    static long[] fac;
    static void populateFact(int n)
    {
        fac = new long[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % MOD;
    }

    static long calculate(int[] arr, int n, int k, int i)
    {
        if(i==n)
        {
            long xor = arr[0], and=arr[0];
            for(int x=1;x<n;x++)
            {
                xor^=arr[x];
                and&=arr[x];
            }
            if(and>=xor)
            {
                return 1;
            }
            return 0;
        }
        long count = 0;
        for(int j=0;j<(1<<k);j++)
        {
            arr[i]=j;
            count+=calculate(arr,n,k,i+1);
        }
        return count;
    }
    static long calc(int n, int k)
    {
        int[] arr = new int[n];
        return calculate(arr,n,k,0);
    }

    public static void main(String[] args) {
        StringBuilder ans = new StringBuilder();
//        int t = ri();
        int t = 1;
        while (t-- > 0)
        {
            int n=ri();
            int m=ri();

            char[][] arr=new char[n][m];

            for(int i=0;i<n;i++)
            {
                arr[i]=rs().toCharArray();
            }

            int[] count = new int[m];
            for(int j=1;j<m;j++)
            {
                int c = 0;
                for(int i=n-1;i>0;i--)
                {
                    if(arr[i][j-1]=='X' && arr[i-1][j]=='X')
                    {
                        c++;
                    }
                }
                count[j] = count[j-1]+c;
            }


            int q=ri();
            for(int i=0;i<q;i++)
            {
                int a=ri()-1;
                int b=ri()-1;
                int sum = count[b]-count[a];

                if(sum>0)
                {
                    ans.append("NO\n");
                }
                else
                {
                    ans.append("YES\n");
                }
            }

        }
        out.print(ans.toString());
        out.flush();
    }

}
/*
4 3
2 1
1 3
3 4
8
3
1 2 3
2 3 1
3
3
2 1 2
2 1 3
3
 */