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 N = 5050, b = 53, mod = 1073676287;
int n, q, pre[N], suf[N], ans[N], binv[N], dp[N][N];
int mul(int x, int y) { return (x * 1ll * y) % mod; }
int add(int x, int y) { return ((x + y) % mod + mod) % mod; }
int pw(int b, int p) {
  if (!p) return 1;
  int ret = pw(b, p / 2);
  ret = mul(ret, ret);
  if (p % 2) ret = mul(ret, b);
  return ret;
}
int getl(int l, int r) {
  int f = add(pre[r], -pre[l - 1]);
  return mul(f, binv[l]);
}
int getr(int l, int r) {
  int f = add(suf[l], -suf[r + 1]);
  return mul(f, binv[n - r + 1]);
}
int check(int l, int r) { return getl(l, r) == getr(l, r); }
int solve(int i, int j) {
  if (i > n || j <= 0 || j < i) return 0;
  int &x = dp[i][j];
  if (x != -1) return x;
  return x = check(i, j) + solve(i, j - 1) + solve(i + 1, j) -
             solve(i + 1, j - 1);
}
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  memset(dp, -1, sizeof dp);
  n = s.size();
  s = "@" + s;
  int bb = 1;
  for (int i = 1; i <= n; i++) {
    binv[i] = pw(bb, mod - 2);
    pre[i] = add(pre[i - 1], mul((s[i] - 'a'), bb));
    bb = mul(bb, b);
  }
  bb = 1;
  for (int i = n; i >= 1; i--) {
    suf[i] = add(suf[i + 1], mul((s[i] - 'a'), bb));
    bb = mul(bb, b);
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << solve(l, r) << '\n';
  }
}
