You've got a string s = s1s2... s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li... ri], which are palindromes.

String s[l... r] = slsl + 1... sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2... s|s|.

String t is called a palindrome, if it reads the same from left to right and from right to left. Formally, if t = t1t2... t|t| = t|t|t|t| - 1... t1.

Input

The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.

It is guaranteed that the given string consists only of lowercase English letters.

Output

Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

Examples

Input

caaaba
5
1 1
1 4
2 3
4 6
4 5


Output

1
7
3
4
2

Note

Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000 + 10;
string s;
int n, q;
bool ok[maxn][maxn];
int l[maxn][maxn];
void update(int u, int v) {
  int y = v;
  int x = u;
  while (x <= n) {
    y = v;
    while (y <= n) {
      l[x][y] = l[x][y] + 1;
      y += y & -y;
    }
    x += x & -x;
  }
}
int get(int u, int v) {
  int r = 0;
  int x = u;
  int y = v;
  while (x > 0) {
    y = v;
    while (y > 0) {
      r = r + l[x][y];
      y -= y & -y;
    }
    x -= x & -x;
  }
  return r;
}
int ans(int x1, int y1, int x2, int y2) {
  return get(x2, y2) - get(x2, y1 - 1) - get(x1 - 1, y2) + get(x1 - 1, y1 - 1);
}
int main() {
  cin >> s;
  n = s.length();
  s = '#' + s;
  memset(ok, false, sizeof(ok));
  for (int i = 1; i <= n; i++) {
    ok[i][i] = true;
    if (i > 1) ok[i][i - 1] = true;
  }
  for (int i = n - 1; i >= 1; i--)
    for (int j = i + 1; j <= n; j++) {
      if (s[i] == s[j] && ok[i + 1][j - 1]) {
        ok[i][j] = true;
      }
    }
  memset(l, 0, sizeof(l));
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      if (ok[i][j]) {
        update(i, j);
      }
  scanf("%d", &q);
  int j, u, v, res;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &u, &v);
    printf("%d\n", ans(u, u, v, v));
  }
  return 0;
}
