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;
char s[5005];
int dp[5005][5005];
bool isPalin[5005][5005];
int n;
void clearmem() {
  memset(dp, 0, sizeof dp);
  memset(isPalin, 0, sizeof isPalin);
}
void preCompute() {
  clearmem();
  for (int i = 0; i < n; i++) {
    isPalin[i][i] = true;
    dp[i][i] = 1;
    if (i < n - 1 && s[i] == s[i + 1]) isPalin[i][i + 1] = true;
  }
  for (int sz = 3; sz <= n; sz++) {
    for (int i = 0; i < n - sz + 1; i++) {
      isPalin[i][i + sz - 1] =
          (isPalin[i + 1][i + sz - 2] && (s[i] == s[i + sz - 1]));
    }
  }
}
void compute() {
  preCompute();
  for (int i = 0; i < n - 1; i++) {
    if (isPalin[i][i + 1])
      dp[i][i + 1] = 1;
    else
      dp[i][i + 1] = 0;
    dp[i][i + 1] += dp[i][i] + dp[i + 1][i + 1];
  }
  for (int sz = 3; sz <= n; sz++) {
    for (int i = 0; i < n - sz + 1; i++) {
      if (isPalin[i][i + sz - 1])
        dp[i][i + sz - 1] = 1;
      else
        dp[i][i + sz - 1] = 0;
      dp[i][i + sz - 1] +=
          dp[i + 1][i + sz - 1] + dp[i][i + sz - 2] - dp[i + 1][i + sz - 2];
    }
  }
}
int main() {
  int q;
  scanf("%s", s);
  scanf("%d", &q);
  n = strlen(s);
  compute();
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", dp[l - 1][r - 1]);
  }
  return 0;
}
