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 = int(1e5) + 123;
const int inf = 1 << 30;
const int mod = int(1e9) + 7;
const int base = 257;
char second[5005];
int n, m, i, j, len, l, r;
long long d[5003][5003];
int hl[5005], hr[5005], pw[5005];
int getL(int l, int r) {
  return (hl[r] + mod - (1ll * hl[l - 1] * pw[r - l + 1]) % mod) % mod;
}
int getR(int l, int r) {
  return (hr[l] + mod - (1ll * hr[r + 1] * pw[r - l + 1]) % mod) % mod;
}
int main() {
  gets(second);
  n = strlen(second);
  pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    hl[i] = (1ll * hl[i - 1] * base + second[i - 1]) % mod;
    pw[i] = (1ll * pw[i - 1] * base) % mod;
    d[i][i] = 1;
  }
  for (int i = n; i >= 1; i--) {
    hr[i] = (1ll * hr[i + 1] * base + second[i - 1]) % mod;
  }
  for (len = 2; len <= n; len++)
    for (i = 1; i + len - 1 <= n; i++) {
      j = i + len - 1;
      d[i][j] = d[i + 1][j] + d[i][j - 1] - d[i + 1][j - 1];
      if (getL(i, j) == getR(i, j)) d[i][j]++;
    }
  scanf("\n%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("\n%d %d", &l, &r);
    printf("%I64d\n", d[l][r]);
  }
  return 0;
}
