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;
int n;
int p;
class cp {
 public:
  bool operator()(pair<long long, long long> p1,
                  pair<long long, long long> p2) {
    if (p1.first / p < p2.first / p)
      return true;
    else if (p1.first / p == p2.first / p) {
      if (p1.first == p2.first && p1.second == p2.second) return false;
      if (p1.second <= p2.second) return true;
      return false;
    }
    return false;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q, i, i1, j1, j, k, l, t, m, temp = 0, ans = 0;
  string s;
  cin >> s;
  n = s.size();
  p = ceil(sqrt(n));
  set<pair<long long, long long>, cp> st;
  map<pair<long long, long long>, long long> mp;
  cin >> q;
  vector<pair<long long, long long>> has(q);
  for (i = 0; i <= q - 1; i++) {
    cin >> j >> k;
    j--;
    k--;
    st.insert(make_pair(j, k));
    has[i] = make_pair(j, k);
  }
  ans = 0;
  long long left, right;
  left = right = 0;
  for (auto pos = st.begin(); pos != st.end(); ++pos) {
    i = pos->first;
    j = pos->second;
    while (j > right) {
      right++;
      if (s[right] == s[right - 1]) ans++;
    }
    while (right > j) {
      if (s[right] == s[right - 1]) ans--;
      right--;
    }
    while (left < i) {
      if (s[left] == s[left + 1]) ans--;
      left++;
    }
    while (left > i) {
      left--;
      if (s[left] == s[left + 1]) ans++;
    }
    mp[make_pair(i, j)] = ans;
  }
  for (i = 0; i < q; i++) {
    cout << mp[make_pair(has[i].first, has[i].second)] << endl;
  }
  return 0;
}