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;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int gcd(int a, int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline int And(int mask, int bit) { return mask & (1 << bit); }
inline int Or(int mask, int bit) { return mask | (1 << bit); }
inline int Xor(int mask, int bit) { return mask & (~(1 << bit)); }
char s[100100];
int n, m, x, y;
int tree[8 * 100100], a[100100];
void build(int pos, int l, int r) {
  if (l == r) {
    tree[pos] = a[l];
    return;
  }
  int mid = (l + r) >> 1, c = 2 * pos;
  build(c, l, mid);
  build(c + 1, mid + 1, r);
  tree[pos] = tree[c] + tree[c + 1];
}
int getSum(int pos, int l, int r, int x, int y) {
  if (r < x || l > y) return 0;
  if (x <= l && r <= y) return tree[pos];
  int mid = (l + r) >> 1, c = 2 * pos;
  return getSum(c, l, mid, x, y) + getSum(c + 1, mid + 1, r, x, y);
}
int main() {
  gets(s);
  n = strlen(s);
  a[1] = 0;
  for (int i = 1; i <= n - 1; i++)
    if (s[i - 1] == s[i])
      a[i + 1] = 1;
    else
      a[i + 1] = 0;
  build(1, 1, n);
  scanf(" %d ", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    x++;
    printf("%d\n", getSum(1, 1, n, x, y));
  }
}