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 = 5003;
const long long mod = 1e9 + 7;
const long long P = 17;
char s[MAXN];
int sl;
long long phash[MAXN];
long long rphash[MAXN];
long long powP[MAXN], invP[MAXN];
long long my_pow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret *= x;
    ret %= mod;
    x *= x;
    x %= mod;
    n >>= 1;
  }
  return ret;
}
long long mod_inv(long long x) { return my_pow(x, mod - 2); }
void init() {
  long long sum = 0;
  long long tp = 1;
  for (int i = 1; i <= sl; i++) {
    phash[i] = (phash[i - 1] + tp * s[i]) % mod;
    tp *= P;
    tp %= mod;
  }
  tp = 1;
  for (int i = sl; i >= 1; i--) {
    rphash[i] = (rphash[i + 1] + tp * s[i]) % mod;
    tp *= P;
    tp %= mod;
  }
  powP[0] = 1;
  invP[0] = mod_inv(powP[0]);
  for (int i = 1; i < sl; i++) {
    powP[i] = powP[i - 1] * P;
    powP[i] %= mod;
    invP[i] = mod_inv(powP[i]);
  }
}
long long shash(int L, int R) {
  return (phash[R] - phash[L - 1] + mod) % mod * invP[L - 1] % mod;
}
long long rshash(int L, int R) {
  return (rphash[L] - rphash[R + 1] + mod) % mod * invP[sl - R] % mod;
}
int dp[MAXN][MAXN];
int A[MAXN][MAXN];
int main() {
  scanf("%s", s + 1);
  sl = strlen(s + 1);
  init();
  for (int len = 2; len <= sl; len++) {
    for (int i = 1; i + len - 1 <= sl; i++) {
      int mid = (i + (i + len - 1)) / 2;
      if (len % 2 == 0) {
        long long sh = shash(i, mid);
        long long rsh = rshash(mid + 1, i + len - 1);
        if (sh == rsh) {
          A[i][i + len - 1]++;
        }
      } else {
        long long sh = shash(i, mid);
        long long rsh = rshash(mid, i + len - 1);
        if (sh == rsh) {
          A[i][i + len - 1]++;
        }
      }
    }
  }
  for (int i = 1; i <= sl; i++) dp[i][i] = 1;
  for (int len = 2; len <= sl; len++) {
    for (int st = 1; st + len - 1 <= sl; st++) {
      dp[st][st + len - 1] = dp[st][st + len - 2] + dp[st + 1][st + len - 1] -
                             dp[st + 1][st + len - 2] + A[st][st + len - 1];
    }
  }
  int q, l, r;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &l, &r);
    printf("%d\n", dp[l][r]);
  }
  return 0;
}
