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;
long long int _xor(long long int a, long long int b) { return a ^ b; }
long long int _and(long long int a, long long int b) { return a & b; }
long long int _or(long long int a, long long int b) { return a | b; }
long long int _not(long long int a) { return ~a; }
long long int ai[1000], pi[2005];
long long int dp[100001];
int main() {
  long long int b, c = 0, x, y, z, t, i, j, k, l1, l2, m, n, c1 = 0, c2 = 0, q,
                   f = 0, ans = 0;
  string str;
  cin >> str;
  dp[0] = 0;
  for (i = 1; i < str.length(); i++) {
    if (str[i] == str[i - 1]) {
      dp[i] = dp[i - 1] + 1;
    } else
      dp[i] = dp[i - 1];
  }
  scanf("%lld", &m);
  long long int low, up;
  for (i = 0; i < m; i++) {
    scanf("%lld", &low);
    scanf("%lld", &up);
    up--;
    low--;
    cout << dp[up] - dp[low] << '\n';
  }
  return 0;
}