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 = 5007;
char s[maxn];
struct PAM {
  int next[maxn][26];
  int fail[maxn];
  int cnt[maxn];
  int num[maxn];
  int len[maxn];
  int S[maxn];
  int last, n, sz;
  int newnode(int l) {
    for (int i = 0; i < 26; i++) {
      next[sz][i] = 0;
    }
    cnt[sz] = 0;
    len[sz] = l;
    return sz++;
  }
  void init() {
    sz = 0;
    newnode(0);
    newnode(-1);
    last = 0;
    n = 0;
    S[n] = -1;
    fail[0] = 1;
  }
  int getfail(int x) {
    while (S[n - len[x] - 1] != S[n]) x = fail[x];
    return x;
  }
  void insert(char ch) {
    int c = ch - 'a';
    S[++n] = c;
    int u = getfail(last);
    if (!next[u][c]) {
      int v = newnode(len[u] + 2);
      fail[v] = next[getfail(fail[u])][c];
      next[u][c] = v;
      num[v] = num[fail[v]] + 1;
    }
    last = next[u][c];
    cnt[last]++;
  }
  void count() {
    for (int i = sz - 1; i; i--) {
      cnt[fail[i]] += cnt[i];
    }
  }
} pam;
int n;
int ans[5003][5003];
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int l = 0; l < n; l++) {
    pam.init();
    int sum = 0;
    for (int r = l; r < n; r++) {
      pam.insert(s[r]);
      sum += pam.num[pam.last];
      ans[l][r] = sum;
    }
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", ans[l - 1][r - 1]);
  }
}
