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;
int di[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dj[] = {1, -1, 0, 0, -1, 1, 1, -1};
const int N = 5e3 + 5, MOD = 1e9 + 7;
int n;
int dp[N][N];
int hsh[N], rhsh[N], power[N];
int mul(int a, int b) {
  int ret = (a * 1ll * b) % MOD;
  while (ret < 0) ret += MOD;
  while (ret >= MOD) ret -= MOD;
  return ret;
}
int add(int a, int b) {
  a += b;
  while (a < 0) a += MOD;
  while (a >= MOD) a -= MOD;
  return a;
}
int get(int l, int r) {
  int x = max(0, (n - r) - (l - 1));
  return mul(add(hsh[r], -hsh[l - 1]), power[x]);
}
int Rget(int l, int r) {
  int x = max(0, (l - 1) - (n - r));
  return mul(add(rhsh[l], -rhsh[r + 1]), power[x]);
}
bool pal(int l, int r) { return get(l, r) == Rget(l, r); }
int main() {
  string s;
  cin >> s;
  for (char &c : s) c = c - 'a' + 1;
  n = s.size();
  s = ' ' + s;
  power[0] = 1;
  for (int i = 1; i <= n; i++) {
    power[i] = mul(power[i - 1], 37);
    hsh[i] = add(hsh[i - 1], mul(s[i], power[i]));
    dp[i][i] = 1;
  }
  for (int i = n; i >= 1; i--)
    rhsh[i] = add(rhsh[i + 1], mul(s[i], power[n - i + 1]));
  for (int len = 2; len <= n; len++)
    for (int i = 1; i + len - 1 <= n; i++)
      dp[i][i + len - 1] = dp[i + 1][i + len - 1] + dp[i][i + len - 2] -
                           dp[i + 1][i + len - 2] + pal(i, i + len - 1);
  int q, l, r;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    printf("%d\n", dp[l][r]);
  }
}
