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;
const long long mod = 1e9 + 7;
const int Max = 1e5 + 5;
const long long IFN = 1e18 + 5;
inline void norm(long long &a) {
  a %= mod;
  (a < 0) && (a += mod);
}
inline long long modAdd(long long a, long long b) {
  a %= mod, b %= mod;
  norm(a), norm(b);
  return (a + b) % mod;
}
inline long long modSub(long long a, long long b) {
  a %= mod, b %= mod;
  norm(a), norm(b);
  return (a - b) % mod;
}
inline long long modMul(long long a, long long b) {
  a %= mod, b %= mod;
  norm(a), norm(b);
  return (a * b) % mod;
}
void F_I_O() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  int t = 1;
  while (t--) {
    string s;
    cin >> s;
    int len = s.size();
    vector<int> v(len + 1, 0);
    int cnt = 0;
    for (int i = 1; i < s.size(); i++) {
      if (s[i] == s[i - 1]) {
        cnt++;
      }
      v[i] = cnt;
    }
    int n;
    int x, y;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> x >> y;
      x--;
      y--;
      cout << v[y] - v[x] << '\n';
    }
  }
}