#include <bits/stdc++.h>
using namespace std;
const int N = 111111;
char ch[N];
int sum[N];
int main() {
  int n, l, r, len;
  while (cin >> ch + 1) {
    scanf("%d", &n);
    sum[0] = 0;
    len = strlen(ch + 1);
    for (int i = 1; i < len; i++)
      if (ch[i] == ch[i + 1])
        sum[i] = sum[i - 1] + 1;
      else
        sum[i] = sum[i - 1];
    while (n--) {
      scanf("%d%d", &l, &r);
      printf("%d\n", sum[r - 1] - sum[l - 1]);
    }
  }
}

### 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>
using namespace std;
const int N = 111111;
char ch[N];
int sum[N];
int main() {
  int n, l, r, len;
  while (cin >> ch + 1) {
    scanf("%d", &n);
    sum[0] = 0;
    len = strlen(ch + 1);
    for (int i = 1; i < len; i++)
      if (ch[i] == ch[i + 1])
        sum[i] = sum[i - 1] + 1;
      else
        sum[i] = sum[i - 1];
    while (n--) {
      scanf("%d%d", &l, &r);
      printf("%d\n", sum[r - 1] - sum[l - 1]);
    }
  }
}

```