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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int IN = 2e5 + 7;
int sum[IN];
int main() {
  string str;
  cin >> str;
  int n;
  scanf("%d", &n);
  memset(sum, 0, sizeof(sum));
  if (str[0] == str[1])
    sum[0] = 1;
  else
    sum[0] = 0;
  int k = str.size();
  for (int i = 1; i < k - 1; i++) {
    if (str[i] == str[i + 1]) {
      sum[i] = sum[i - 1] + 1;
    } else {
      sum[i] = sum[i - 1];
    }
  }
  sum[k - 1] = sum[k - 2];
  int l, r;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &l, &r);
    if (l != 1)
      printf("%d\n", sum[r - 2] - sum[l - 2]);
    else
      printf("%d\n", sum[r - 2]);
  }
}