#include <bits/stdc++.h>
using namespace std;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
pair<int, int> M[8] = {make_pair(0, 1),  make_pair(1, 0),  make_pair(-1, 0),
                       make_pair(0, -1), make_pair(-1, 1), make_pair(-1, -1),
                       make_pair(1, -1), make_pair(1, 1)};
string s;
int val[100005];
int main() {
  cin >> s;
  int l = s.length();
  for (int(a) = (0); (a) <= (l - 1); (a)++) {
    if (s[a] == s[a - 1]) val[a] = 1;
  }
  for (int(a) = (0); (a) <= (l - 1); (a)++) {
    val[a] += val[a - 1];
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", val[r - 1] - val[l - 1]);
  }
}

### Prompt

Please create a solution in Cpp 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;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
pair<int, int> M[8] = {make_pair(0, 1),  make_pair(1, 0),  make_pair(-1, 0),
                       make_pair(0, -1), make_pair(-1, 1), make_pair(-1, -1),
                       make_pair(1, -1), make_pair(1, 1)};
string s;
int val[100005];
int main() {
  cin >> s;
  int l = s.length();
  for (int(a) = (0); (a) <= (l - 1); (a)++) {
    if (s[a] == s[a - 1]) val[a] = 1;
  }
  for (int(a) = (0); (a) <= (l - 1); (a)++) {
    val[a] += val[a - 1];
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", val[r - 1] - val[l - 1]);
  }
}

```