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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void fastInOut() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
long long pw(long long b, long long p) {
  if (!p) return 1;
  long long sq = pw(b, p / 2);
  sq *= sq;
  if (p % 2) sq *= b;
  return sq;
}
int sd(long long x) { return x < 10 ? x : x % 10 + sd(x / 10); }
int n, m;
string str1, str2;
char C;
int main() {
  fastInOut();
  string s;
  cin >> s;
  cin >> n;
  int arr[1000000];
  arr[0] = 0;
  for (int i = 1; i <= ((int)((s).size())); i++) {
    if (s[i] == s[i - 1])
      arr[i + 1] = arr[i] + 1;
    else
      arr[i + 1] = arr[i];
  }
  for (long long i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    cout << arr[b] - arr[a] << endl;
  }
  return 0;
}