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;
using ll = long long;
using ld = long double;
using ul = unsigned long long;
using ii = pair<int, int>;
const ll mod7 = 1000000007, mod = 1000000000, OO = (1LL << 62);
const ld EPS = 1e-7;
double Pi = acos(-1);
const ld rad = Pi / 180.0;
ll power(ll x, ll y, ll p) {
  ll res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
ll nCr(ll n, ll m) { return (m == 0) ? 1 : n * nCr(n - 1, m - 1) / m; }
int main() {
  string s;
  cin >> s;
  int m, len;
  len = (int)s.length();
  int a[len], pf[len];
  cin >> m;
  a[0] = 0;
  for (int i = 1; i < len; ++i)
    if (s[i] == s[i - 1])
      a[i] = 1;
    else
      a[i] = 0;
  for (int i = 0; i < len; ++i) {
    if (i == 0) {
      pf[i] = a[i];
    } else {
      pf[i] = pf[i - 1] + a[i];
    }
  }
  while (m--) {
    int l, r;
    cin >> l >> r;
    cout << pf[r - 1] - pf[l - 1] << "\n";
  }
  return 0;
}