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>
int main() {
  char str[100007];
  int num_quer;
  int left, right;
  int dp_n[100007];
  int dp_sum[100007];
  scanf("%s", str);
  scanf("%d", &num_quer);
  memset(dp_n, 0, sizeof(dp_n));
  memset(dp_sum, 0, sizeof(dp_sum));
  int len = strlen(str);
  dp_n[0] = dp_sum[0] = 0;
  for (int i = 1; i < len; i++) {
    if (str[i - 1] == str[i]) {
      dp_n[i] = 1;
      dp_sum[i] = dp_sum[i - 1] + 1;
    } else {
      dp_n[i] = 0;
      dp_sum[i] = dp_sum[i - 1];
    }
  }
  while (num_quer > 0) {
    scanf("%d %d", &left, &right);
    if (num_quer > 1)
      printf("%d\n", (dp_sum[right - 1] - dp_sum[left - 1]));
    else
      printf("%d", (dp_sum[right - 1] - dp_sum[left - 1]));
    num_quer--;
  }
}