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;
template <typename T>
void print(vector<T> v) {
  for (T &i : v) cout << i << " ";
  cout << "\n";
}
template <typename T>
void read(vector<T> &v) {
  for (T &i : v) cin >> i;
}
long long mod = 1e9 + 7;
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  int m;
  cin >> m;
  vector<int> v(n);
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) {
      v[i] = 1;
    } else {
      v[i] = 0;
    }
  }
  for (int i = 1; i < n - 1; i++) {
    v[i] += v[i - 1];
  }
  v[n - 1] = v[n - 2];
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    if (x == 0)
      cout << v[y - 1] << "\n";
    else
      cout << v[y - 1] - v[x - 1] << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int tcs = 1;
  while (tcs--) solve();
  return 0;
}