#include <bits/stdc++.h>
using namespace std;
long long knitX[] = {+2, +2, -2, -2, +1, +1, -1, -1};
long long knitY[] = {+1, -1, +1, -1, +2, -2, +2, -2};
int main() {
  char s[1000007];
  while (scanf("%s", s) == 1) {
    int l = strlen(s);
    int dp[1000007];
    dp[1] = 0;
    for (int i = 2; i <= l; i++) {
      if (s[i - 1] == s[i - 2])
        dp[i] = dp[i - 1] + 1;
      else
        dp[i] = dp[i - 1];
    }
    int T, x, y;
    scanf("%d", &T);
    for (int tt = 1; tt <= T; tt++) {
      scanf("%d %d", &x, &y);
      printf("%d\n", dp[y] - dp[x]);
    }
  }
  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;
long long knitX[] = {+2, +2, -2, -2, +1, +1, -1, -1};
long long knitY[] = {+1, -1, +1, -1, +2, -2, +2, -2};
int main() {
  char s[1000007];
  while (scanf("%s", s) == 1) {
    int l = strlen(s);
    int dp[1000007];
    dp[1] = 0;
    for (int i = 2; i <= l; i++) {
      if (s[i - 1] == s[i - 2])
        dp[i] = dp[i - 1] + 1;
      else
        dp[i] = dp[i - 1];
    }
    int T, x, y;
    scanf("%d", &T);
    for (int tt = 1; tt <= T; tt++) {
      scanf("%d %d", &x, &y);
      printf("%d\n", dp[y] - dp[x]);
    }
  }
  return 0;
}

```