#include <bits/stdc++.h>
using namespace std;
const int INF_int = 100000000;
const long long INF_ll = 1000000000000000;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string a;
  cin >> a;
  int n = a.size();
  int m;
  cin >> m;
  int b[n];
  for (int i = 0; i < n; i++) {
    b[i] = 0;
    if (i != n - 1 && a[i] == a[i + 1]) {
      b[i] = 1;
    }
  }
  int sum[n];
  sum[0] = b[0];
  for (int i = 1; i < n; i++) {
    sum[i] = b[i] + sum[i - 1];
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    if (x != 0)
      cout << sum[y - 1] - sum[x - 1] << "\n";
    else
      cout << sum[y - 1] << "\n";
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
const int INF_int = 100000000;
const long long INF_ll = 1000000000000000;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string a;
  cin >> a;
  int n = a.size();
  int m;
  cin >> m;
  int b[n];
  for (int i = 0; i < n; i++) {
    b[i] = 0;
    if (i != n - 1 && a[i] == a[i + 1]) {
      b[i] = 1;
    }
  }
  int sum[n];
  sum[0] = b[0];
  for (int i = 1; i < n; i++) {
    sum[i] = b[i] + sum[i - 1];
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    if (x != 0)
      cout << sum[y - 1] - sum[x - 1] << "\n";
    else
      cout << sum[y - 1] << "\n";
  }
}

```