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() {
  int L, D, i, j, N, dp[100000] = {0};
  char arr[100001];
  scanf("%s", arr);
  for (i = 0; i < strlen(arr) - 1; i++) {
    if (i == 0 && arr[i] == arr[i + 1])
      dp[0] = 1;
    else if (arr[i] == arr[i + 1])
      dp[i] = dp[i - 1] + 1;
    else if (i != 0)
      dp[i] = dp[i - 1];
  }
  dp[i] = dp[i - 1];
  arr[i + 1] = 'a';
  scanf("%d", &N);
  int sol;
  for (i = 0; i < N; i++) {
    scanf("%d %d", &L, &D);
    if (arr[L - 1] == arr[L] && arr[D - 1] == arr[D])
      sol = dp[D - 1] - dp[L - 1];
    else if (arr[L - 1] == arr[L] && arr[D - 1] != arr[D])
      sol = dp[D - 1] - dp[L - 1] + 1;
    else if (arr[L - 1] != arr[L] && arr[D - 1] == arr[D])
      sol = dp[D - 1] - dp[L - 1] - 1;
    else if (arr[L - 1] != arr[L] && arr[D - 1] != arr[D])
      sol = dp[D - 1] - dp[L - 1];
    printf("%d\n", sol);
  }
  return 0;
}