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 long long inf = (long long)1e9 + 10;
const long long linf = (long long)1e18 + 10;
const long long mod = (long long)(1e9 + 7);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};
const double eps = 1e-10;
long long mop(long long a, long long b, long long m = mod) {
  long long r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool ool(int x, int y, int h, int w) {
  return ((x < 0) || (h <= x) || (y < 0) || (w <= y));
}
bool deq(double a, double b) { return abs(a - b) < eps; }
struct oreno_initializer {
  oreno_initializer() {
    cin.tie(0);
    ios::sync_with_stdio(0);
  }
} oreno_initializer;
string s;
int n, dp[5050][5050], dp2[5050][5050], q, l, r;
bool pa[5050][5050];
signed main() {
  cin >> s;
  n = s.size();
  for (int(i) = (0); (i) < (n); ++(i)) pa[i + 1][i + 1] = 1;
  for (int(k) = (1); (k) < (n); ++(k))
    for (int(i) = (0); (i) < (n); ++(i))
      if (i + k < n) {
        int j = i + k;
        if (k == 1)
          pa[i + 1][j + 1] = (s[i] == s[j]);
        else
          pa[i + 1][j + 1] = (pa[i + 2][j] && s[i] == s[j]);
      }
  for (int(i) = (0); (i) < (n); ++(i)) dp2[i + 1][i + 1] = 1;
  for (int j = n; j > 0; j--)
    for (int i = j - 1; i > 0; i--)
      dp2[i][j] = (dp2[i + 1][j] + pa[i][j]) % mod;
  for (int(i) = (0); (i) < (n); ++(i)) dp[i + 1][i + 1] = 1;
  for (int(i) = (1); (i) < (n + 1); ++(i))
    for (int(j) = (i + 1); (j) < (n + 1); ++(j))
      dp[i][j] = (dp[i][j - 1] + dp2[i][j]) % mod;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    printf("%d\n", dp[l][r]);
  }
}
