#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
const long long int N = 1e6 + 10;
const long long int M = 1e12 + 10;
long long int pf[N];
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int sumofD(long long int a) {
  long long int sum = 0;
  while (a) {
    sum += a % 10;
    a /= 10;
  }
  return sum;
}
long long int findGCD(vector<long long int> arr) {
  long long int result = arr[0];
  for (long long int i = 1; i < arr.size(); i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
void printArr(vector<long long int> arr) {
  for (auto &i : arr) {
    cout << i << " ";
  }
  cout << endl;
}
bool perfectsquare(long long int num) {
  long long int sin = sqrt(num);
  if (sin * sin == num) {
    return true;
  }
  return false;
}
bool binsearch(long long int arr[], long long int s, long long int key) {
  long long int l = 0, r = s - 1;
  while (r >= l) {
    long long int mid = (r + l) / 2;
    if (arr[mid] == key)
      return true;
    else if (arr[mid] > key)
      r = mid - 1;
    else if (arr[mid] < key)
      l = mid + 1;
  }
  return false;
}
void solve() {
  string in;
  cin >> in;
  int n = in.size();
  for (int i = 1; i < n; i++) {
    if (in[i - 1] == in[i]) pf[i]++;
  }
  for (int i = 1; i <= n; i++) pf[i] += pf[i - 1];
  int m;
  cin >> m;
  while (m--) {
    long long int l, r;
    cin >> l >> r;
    cout << pf[r - 1] - pf[l - 1] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
const long long int N = 1e6 + 10;
const long long int M = 1e12 + 10;
long long int pf[N];
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int sumofD(long long int a) {
  long long int sum = 0;
  while (a) {
    sum += a % 10;
    a /= 10;
  }
  return sum;
}
long long int findGCD(vector<long long int> arr) {
  long long int result = arr[0];
  for (long long int i = 1; i < arr.size(); i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
void printArr(vector<long long int> arr) {
  for (auto &i : arr) {
    cout << i << " ";
  }
  cout << endl;
}
bool perfectsquare(long long int num) {
  long long int sin = sqrt(num);
  if (sin * sin == num) {
    return true;
  }
  return false;
}
bool binsearch(long long int arr[], long long int s, long long int key) {
  long long int l = 0, r = s - 1;
  while (r >= l) {
    long long int mid = (r + l) / 2;
    if (arr[mid] == key)
      return true;
    else if (arr[mid] > key)
      r = mid - 1;
    else if (arr[mid] < key)
      l = mid + 1;
  }
  return false;
}
void solve() {
  string in;
  cin >> in;
  int n = in.size();
  for (int i = 1; i < n; i++) {
    if (in[i - 1] == in[i]) pf[i]++;
  }
  for (int i = 1; i <= n; i++) pf[i] += pf[i - 1];
  int m;
  cin >> m;
  while (m--) {
    long long int l, r;
    cin >> l >> r;
    cout << pf[r - 1] - pf[l - 1] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

```