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>
char s[10000];
int n;
const long long mod = 1000000007;
const long long x = 143523455;
long long lhash[10000];
long long rhash[10000];
long long powx[10000];
int dp[5050][5050];
int get_dp(int a, int b) {
  if (dp[a][b] != -1) return dp[a][b];
  if (a == b) {
    dp[a][b] = 1;
    return dp[a][b];
  }
  if (a + 1 == b) {
    dp[a][b] = 2 + ((int)s[a - 1] == s[b - 1]);
    return dp[a][b];
  }
  int sum = get_dp(a + 1, b) + get_dp(a, b - 1) - get_dp(a + 1, b - 1);
  long long h1 = lhash[b] - lhash[a - 1] * powx[b - a + 1];
  h1 %= mod;
  h1 += mod;
  h1 %= mod;
  long long h2 = rhash[n - a + 1] - rhash[n - b] * powx[b - a + 1];
  h2 %= mod;
  h2 += mod;
  h2 %= mod;
  if (h1 == h2) {
    sum++;
  }
  dp[a][b] = sum;
  return dp[a][b];
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  lhash[0] = 0;
  rhash[0] = 0;
  powx[0] = 1;
  for (int i = 1; i <= n; i++) {
    powx[i] = powx[i - 1] * x;
    powx[i] %= mod;
    lhash[i] = lhash[i - 1] * x + (long long)s[i - 1];
    lhash[i] %= mod;
    rhash[i] = rhash[i - 1] * x + (long long)s[n - i];
    rhash[i] %= mod;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = -1;
    }
  }
  get_dp(1, n);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", get_dp(a, b));
  }
}
