#include <bits/stdc++.h>
using namespace std;
char s[100010];
bool check[100010];
int cnt[100010];
int m;
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) check[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (check[i] == 1)
      cnt[i] = cnt[i - 1] + 1;
    else
      cnt[i] = cnt[i - 1];
  }
  scanf("%d", &m);
  for (int t = 0; t < m; t++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    int a, b;
    if (l - 1 < 0)
      a = 0;
    else
      a = cnt[l - 1];
    b = cnt[r - 1];
    printf("%d\n", b - a);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
char s[100010];
bool check[100010];
int cnt[100010];
int m;
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) check[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (check[i] == 1)
      cnt[i] = cnt[i - 1] + 1;
    else
      cnt[i] = cnt[i - 1];
  }
  scanf("%d", &m);
  for (int t = 0; t < m; t++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    int a, b;
    if (l - 1 < 0)
      a = 0;
    else
      a = cnt[l - 1];
    b = cnt[r - 1];
    printf("%d\n", b - a);
  }
  return 0;
}

```