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 power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    long long int t = power(a, b / 2);
    if (b % 2 == 0)
      return ((t % 1000000007) * (t % 1000000007)) % 1000000007;
    else
      return ((((t % 1000000007) * (t % 1000000007)) % 1000000007) *
              ((a % 1000000007) % 1000000007)) %
             1000000007;
  }
}
void solve() {
  long long int n, l, r, a[100000], i;
  string s;
  cin >> s;
  for (i = 0; i < s.length() - 1; i++) {
    if (s[i] == s[i + 1]) {
      a[i] = 1;
    } else
      a[i] = 0;
  }
  a[s.length() - 1] = 0;
  for (i = 1; i < s.length(); i++) {
    a[i] += a[i - 1];
  }
  cin >> n;
  while (n--) {
    cin >> l >> r;
    if (l - 2 < 0)
      cout << a[r - 2] << "\n";
    else
      cout << a[r - 2] - a[l - 2] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}