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 = 5050;
int dp[maxn][maxn];
bool ha[maxn][maxn];
char str[maxn];
void DP(void) {
  int n = strlen(str + 1);
  for (int i = 1; i <= n; i++) dp[i][i] = 1, ha[i][i] = true;
  for (int i = 2; i <= n; i++)
    if (str[i] == str[i - 1]) ha[i - 1][i] = true;
  for (int len = 2; len <= n; len++) {
    for (int l = 1; l <= n - len + 1; l++) {
      int r = l + len - 1;
      if (str[l] == str[r] && ha[l + 1][r - 1]) ha[l][r] = true;
      dp[l][r] = dp[l][r - 1] + dp[l + 1][r] - dp[l + 1][r - 1] + ha[l][r];
    }
  }
  return;
}
int main(void) {
  scanf("%s", str + 1);
  DP();
  int n;
  scanf("%d", &n);
  int x, y;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    printf("%d\n", dp[x][y]);
  }
}
