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>
const int N = 5000 + 2;
bool vis[N][N];
int ans[N][N];
char s[N];
int n;
int main() {
  scanf("%s", s);
  n = strlen(s);
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < n; ++i) {
    vis[i][i] = true;
    for (int j = 0; j < i; ++j) {
      if (j == i - 1)
        vis[j][i] = (s[i] == s[i - 1]);
      else
        vis[j][i] = (s[j] == s[i]) && vis[j + 1][i - 1];
    }
  }
  memset(ans, 0, sizeof ans);
  for (int i = 0; i < n; ++i) {
    ans[i][i] = 1;
    for (int j = i - 1; j >= 0; --j) {
      ans[j][i] +=
          ans[j + 1][i] + ans[j][i - 1] - ans[j + 1][i - 1] + (int)vis[j][i];
    }
  }
  int Q, l, r;
  scanf("%d", &Q);
  while (Q-- > 0) {
    scanf("%d%d", &l, &r);
    --l;
    --r;
    printf("%d\n", ans[l][r]);
  }
  return 0;
}
