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;
struct simplehash {
  int len;
  long long base, mod;
  vector<int> P, H, R;
  simplehash() {}
  simplehash(const char* str, long long b, long long m) {
    base = b, mod = m, len = strlen(str);
    P.resize(len + 3, 1), H.resize(len + 3, 0), R.resize(len + 3, 0);
    for (int i = 1; i <= len; i++) P[i] = (P[i - 1] * base) % mod;
    for (int i = 1; i <= len; i++)
      H[i] = (H[i - 1] * base + str[i - 1] + 1007) % mod;
    for (int i = len; i >= 1; i--)
      R[i] = (R[i + 1] * base + str[i - 1] + 1007) % mod;
  }
  inline int range_hash(int l, int r) {
    int hashval = H[r + 1] - ((long long)P[r - l + 1] * H[l] % mod);
    return (hashval < 0 ? hashval + mod : hashval);
  }
  inline int reverse_hash(int l, int r) {
    ;
    int hashval = R[l + 1] - ((long long)P[r - l + 1] * R[r + 2] % mod);
    return (hashval < 0 ? hashval + mod : hashval);
  }
};
struct stringhash {
  simplehash sh1, sh2;
  stringhash() {}
  stringhash(const char* str) {
    sh1 = simplehash(str, 1949313259, 2091573227);
    sh2 = simplehash(str, 1997293877, 2117566807);
  }
  inline long long range_hash(int l, int r) {
    return ((long long)sh1.range_hash(l, r) << 32) ^ sh2.range_hash(l, r);
  }
  inline long long reverse_hash(int l, int r) {
    return ((long long)sh1.reverse_hash(l, r) << 32) ^ sh2.reverse_hash(l, r);
  }
} S;
char str[5002], ss[5002];
int len;
int sum[5002][5002], row[5002][5002], pp;
inline int getsum(int x1, int y1, int x2, int y2) {
  return sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
}
int main() {
  int l, r, q, i, j;
  scanf("%s", str);
  len = strlen(str);
  ss[0] = '#';
  for (i = 0; i < len; i++) ss[i + 1] = str[i];
  S = stringhash(ss);
  for (i = 1; i <= len; i++) {
    for (j = i; j <= len; j++) {
      if (S.range_hash(i, j) == S.reverse_hash(i, j)) sum[i][j] = 1;
    }
  }
  for (i = 1; i <= len; i++) {
    for (j = 1; j <= len; j++) {
      row[i][j] = row[i][j - 1] + sum[i][j];
    }
  }
  for (i = 0; i <= len; i++)
    for (j = 0; j <= len; j++) sum[i][j] = 0;
  for (i = 1; i <= len; i++) {
    for (j = 1; j <= len; j++) {
      sum[i][j] = sum[i - 1][j] + row[i][j];
    }
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d %d", &l, &r);
    printf("%d\n", getsum(l, l, r, r));
  }
}
