Problem: Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.

You've got string s = s1s2... sn (n is the length of the string), consisting only of characters "." and "#" and m queries. Each query is described by a pair of integers li, ri (1 ≤ li < ri ≤ n). The answer to the query li, ri is the number of such integers i (li ≤ i < ri), that si = si + 1.

Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.

Input

The first line contains string s of length n (2 ≤ n ≤ 105). It is guaranteed that the given string only consists of characters "." and "#".

The next line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains the description of the corresponding query. The i-th line contains integers li, ri (1 ≤ li < ri ≤ n).

Output

Print m integers — the answers to the queries in the order in which they are given in the input.

Examples

Input

......
4
3 4
2 3
1 6
2 6


Output

1
1
5
4


Input

#..###
5
1 3
5 6
1 5
3 6
3 4


Output

1
1
2
2
0

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		Scanner in = new Main().new Scanner(inputStream);
		Main solver = new Main();
		solver.solve(in);
	}

	public void solve(Scanner in) {
		String s = in.readLine();
		int n = s.length(), b = 0;
		int[] c = new int[n+10];
		for(int i = 1; i < n; i++) {
			if(s.charAt(i) == s.charAt(i-1)) {
				c[i+1] = ++b;
			} else {
				c[i+1] = b;
			}
		}
		
		int m = in.ni();
		for(int i = 0; i < m; i++) {
			int l = in.ni();
			int r = in.ni();
			int res = c[r] - c[l];
			if(res < 0) res = 0;
			System.out.println(res);
		}
	}

	class Scanner {
		BufferedReader br;
		StringTokenizer in;

		public Scanner(InputStream inputStream) {
			br = new BufferedReader(new InputStreamReader(inputStream));
		}

		public String next() {
			return hasMoreTokens() ? in.nextToken() : null;
		}

		public String readLine() {
			try {
				return br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
				return null;
			}
		}

		boolean hasMoreTokens() {
			while (in == null || !in.hasMoreTokens()) {
				String s = readLine();
				if (s == null)
					return false;
				in = new StringTokenizer(s);
			}
			return true;
		}

		public String nextString() {
			return next();
		}

		public int ni() {
			return Integer.parseInt(nextString());
		}
		
		public long nl() {
			return Long.parseLong(nextString());
		}

		public int[] nia(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
	}
}