#include <bits/stdc++.h>
using namespace std;
using lli = long long;
const lli OO = 1e18;
void fast() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
lli di[8] = {-1, 1, 0, 0, -1, 1, -1, 1};
lli dj[8] = {0, 0, -1, 1, 1, 1, -1, -1};
lli frq[100001 + 7];
string ys = "YES", no = "NO";
vector<vector<lli>> adj;
vector<lli> vis;
lli mem[101][101][2];
vector<lli> a(100001), b(100001);
int main() {
  string s;
  cin >> s;
  a.resize(s.size());
  a[s.size() - 1] = 0;
  for (int i = s.size() - 2; i >= 0; i--) {
    a[i] = a[i + 1] + (s[i] == s[i + 1]);
  }
  lli q;
  cin >> q;
  while (q--) {
    lli l, r;
    cin >> l >> r;
    l--, r--;
    cout << a[l] - a[r] << "\n";
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using lli = long long;
const lli OO = 1e18;
void fast() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
lli di[8] = {-1, 1, 0, 0, -1, 1, -1, 1};
lli dj[8] = {0, 0, -1, 1, 1, 1, -1, -1};
lli frq[100001 + 7];
string ys = "YES", no = "NO";
vector<vector<lli>> adj;
vector<lli> vis;
lli mem[101][101][2];
vector<lli> a(100001), b(100001);
int main() {
  string s;
  cin >> s;
  a.resize(s.size());
  a[s.size() - 1] = 0;
  for (int i = s.size() - 2; i >= 0; i--) {
    a[i] = a[i + 1] + (s[i] == s[i + 1]);
  }
  lli q;
  cin >> q;
  while (q--) {
    lli l, r;
    cin >> l >> r;
    l--, r--;
    cout << a[l] - a[r] << "\n";
  }
}

```