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;
int gi() {
  int x = 0, w = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') w = 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return w ? x : -x;
}
const int N = 5005;
struct Palindromic_Tree {
  int tr[N][26], fa[N], len[N], dep[N], last, tot;
  void init() {
    memset(tr, 0, sizeof(tr));
    memset(fa, 0, sizeof(fa));
    memset(len, 0, sizeof(len));
    memset(dep, 0, sizeof(dep));
    fa[last = 0] = fa[1] = 1;
    len[tot = 1] = -1;
  }
  void extend(int c, int n, char *s) {
    int v = last;
    while (s[n - len[v] - 1] != s[n]) v = fa[v];
    if (!tr[v][c]) {
      int u = ++tot, k = fa[v];
      len[u] = len[v] + 2;
      while (s[n - len[k] - 1] != s[n]) k = fa[k];
      fa[u] = tr[k][c];
      tr[v][c] = u;
      dep[u] = dep[fa[u]] + 1;
    }
    last = tr[v][c];
  }
} T;
char s[N], ch[N];
int ans[N][N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; ++i) {
    T.init();
    for (int j = i; j <= n; ++j) ch[j - i + 1] = s[j];
    for (int j = i; j <= n; ++j) {
      T.extend(ch[j - i + 1] - 'a', j - i + 1, ch);
      ans[i][j] = ans[i][j - 1] + T.dep[T.last];
    }
  }
  int q = gi();
  while (q--) {
    int l = gi(), r = gi();
    printf("%d\n", ans[l][r]);
  }
  return 0;
}
