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>
int dp[5005][5005] = {0};
int isp[5005][5005] = {0};
char line[5002];
int n, size;
void makePal() {
  for (int i = 1; i <= size; i++) isp[i][i] = 1;
  for (int j = 1; j < size; j++)
    if (line[j] == line[j + 1]) isp[j][j + 1] = 1;
  for (int i = 0; i < size - 2; i++)
    for (int j = 1; j < size - i; j++) {
      int l = j, r = j + i;
      if (isp[l][r] == 1 && (line[l - 1] == line[r + 1])) isp[l - 1][r + 1] = 1;
    }
}
int main() {
  scanf("%s", line);
  size = strlen(line);
  for (int i = size; i >= 1; i--) line[i] = line[i - 1];
  line[0] = ' ';
  scanf("%d", &n);
  makePal();
  for (int i = 1; i <= size; i++)
    for (int j = 1; j <= size; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + isp[i][j];
    }
  while (n--) {
    int init, end;
    scanf("%d %d", &init, &end);
    int res = dp[end][end] - dp[init - 1][end] - dp[end][init - 1] +
              dp[init - 1][init - 1];
    printf("%d\n", res);
  }
  return 0;
}
