#include <bits/stdc++.h>
const int INF = 2e9;
double eps = 1e-9;
int main() {
  std::string str;
  std::cin >> str;
  int m, n = str.length();
  std::vector<int> vec(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    if (str[i] == str[i + 1])
      vec[i + 1] = vec[i] + 1;
    else
      vec[i + 1] = vec[i];
  }
  int r, l, ans;
  std::cin >> m;
  for (; std::cin >> l >> r;) {
    ans = vec[r - 1] - vec[l - 1];
    std::cout << ans << std::endl;
  }
  getchar();
  getchar();
  exit(0);
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
const int INF = 2e9;
double eps = 1e-9;
int main() {
  std::string str;
  std::cin >> str;
  int m, n = str.length();
  std::vector<int> vec(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    if (str[i] == str[i + 1])
      vec[i + 1] = vec[i] + 1;
    else
      vec[i + 1] = vec[i];
  }
  int r, l, ans;
  std::cin >> m;
  for (; std::cin >> l >> r;) {
    ans = vec[r - 1] - vec[l - 1];
    std::cout << ans << std::endl;
  }
  getchar();
  getchar();
  exit(0);
}

```