#include <bits/stdc++.h>
int main() {
  long long int l, m, i, j = 0;
  char str[100001];
  scanf("%s", str);
  l = strlen(str);
  scanf("%lld", &m);
  long long int arr[m][2], arr1[l + 1];
  for (i = 0; i < m; i++) {
    scanf("%lld %lld", &arr[i][0], &arr[i][1]);
  }
  arr1[0] = 0;
  for (i = 0; i < l - 1; i++) {
    arr1[i + 1] = j;
    if (str[i] == str[i + 1]) {
      j++;
    }
  }
  arr1[l] = j;
  for (i = 0; i < m; i++) {
    printf("%lld\n", arr1[arr[i][1]] - arr1[arr[i][0]]);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
int main() {
  long long int l, m, i, j = 0;
  char str[100001];
  scanf("%s", str);
  l = strlen(str);
  scanf("%lld", &m);
  long long int arr[m][2], arr1[l + 1];
  for (i = 0; i < m; i++) {
    scanf("%lld %lld", &arr[i][0], &arr[i][1]);
  }
  arr1[0] = 0;
  for (i = 0; i < l - 1; i++) {
    arr1[i + 1] = j;
    if (str[i] == str[i + 1]) {
      j++;
    }
  }
  arr1[l] = j;
  for (i = 0; i < m; i++) {
    printf("%lld\n", arr1[arr[i][1]] - arr1[arr[i][0]]);
  }
  return 0;
}

```