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 int nax = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INTF = 2147483647;
const long long LLF = 9223372036854775807;
int dp[nax], pfs[nax];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  int n, l, r;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == s[i + 1]) {
      dp[i]++;
    }
  }
  for (int i = 1; i <= s.length(); i++) {
    pfs[i] = pfs[i - 1] + dp[i - 1];
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> r;
    cout << pfs[r - 1] - pfs[l - 1] << "\n";
  }
  return 0;
}