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;
char s[5010];
int n, Dp[5010][5010], T;
unsigned long long iH1[5010], iH2[5010], iPow[5010];
bool ok[5010][5010];
unsigned long long fH1(int l, int r) { return iH1[r] - iH1[l] * iPow[r - l]; }
unsigned long long fH2(int l, int r) { return iH2[l] - iH2[r] * iPow[r - l]; }
bool isGood(int l, int r) { return fH1(l, r + 1) == fH2(l, r + 1); }
void fInit() {
  for (int i = 0; i < n; i++) {
    Dp[i][i] = 1;
    ok[i][i] = ok[i + 1][i] = true;
  }
  iPow[0] = 1;
  for (int i = 1; i <= n; i++) iPow[i] = iPow[i - 1] * 65436534565367ull;
  iH1[0] = 0;
  for (int i = 0; i < n; i++)
    iH1[i + 1] = iH1[i] * 65436534565367ull + (unsigned long long)s[i];
  iH2[n] = 0;
  for (int i = n - 1; i >= 0; i--)
    iH2[i] = iH2[i + 1] * 65436534565367ull + (unsigned long long)s[i];
}
void solve() {
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) ok[i][j] = isGood(i, j);
  for (int len = 2; len <= n; len++)
    for (int i = 0; i + len - 1 < n; i++) {
      int j = i + len - 1;
      Dp[i][j] =
          Dp[i + 1][j] + Dp[i][j - 1] - Dp[i + 1][j - 1] + (int)(ok[i][j]);
    }
}
int main() {
  scanf("%s", &s);
  n = strlen(s);
  fInit();
  solve();
  scanf("%d", &T);
  while (T--) {
    int x, y;
    scanf("%d%d", &x, &y);
    printf("%d\n", Dp[--x][--y]);
  }
  return 0;
}
