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;
bool debug = false;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
vector<int> dp[4][5005];
char s[5005];
void pre() {
  for (int(i) = 0; (i) < (int)(n); (i)++)
    for (int(k) = 0; (k) < (int)(4); (k)++) dp[k][i].resize(n - i + 1);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    dp[0][i][i - i] = 1;
    if (s[i] == s[i + 1]) {
      dp[0][i][i + 1 - i] = 1;
    }
    if (i + 1 < n) {
      dp[1][i][i + 1 - i] = 1;
      dp[2][i][i + 1 - i] = 1;
    }
  }
  for (int l = 3; l <= n; l++) {
    for (int i = 0, j = i + l - 1; j < n; i++, j++) {
      dp[1][i][j - i] = dp[0][i][j - 1 - i] + dp[1][i][j - 1 - i];
      dp[2][i][j - i] = dp[0][i + 1][j - i - 1] + dp[2][i + 1][j - i - 1];
      for (int(k) = 0; (k) < (int)(4); (k)++)
        dp[3][i][j - i] += dp[k][i + 1][j - 1 - i - 1];
      if (s[i] == s[j]) {
        dp[0][i][j - i] += dp[0][i + 1][j - 1 - i - 1];
      }
    }
  }
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  pre();
  scanf("%d", &m);
  for (int i = 0, l, r; i < m; i++) {
    scanf("%d%d", &l, &r);
    l--;
    r--;
    int aa = 0;
    for (int(k) = 0; (k) < (int)(4); (k)++) aa += dp[k][l][r - l];
    printf("%d\n", aa);
  }
  return 0;
}
