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 = 1e6 + 5;
int dp[5005][5005];
struct PAM {
  int nex[maxn][30];
  int fail[maxn];
  int cnt[maxn];
  int num[maxn];
  int len[maxn];
  int S[maxn];
  int last, n, p;
  int create(int rt) {
    memset(nex[p], 0, sizeof(nex[p]));
    cnt[p] = 0;
    num[p] = 0;
    len[p] = rt;
    return p++;
  }
  void init() {
    p = last = n = 0;
    create(0);
    create(-1);
    S[0] = -1;
    fail[0] = 1;
  }
  int getFail(int x) {
    while (S[n - len[x] - 1] != S[n]) x = fail[x];
    return x;
  }
  void insert(int c) {
    c = c - 'a';
    S[++n] = c;
    int cur = getFail(last);
    if (!nex[cur][c]) {
      int now = create(len[cur] + 2);
      fail[now] = nex[getFail(fail[cur])][c];
      nex[cur][c] = now;
      num[now] = num[fail[now]] + 1;
    }
    last = nex[cur][c];
    cnt[last]++;
  }
  void count() {
    long long ans = 0;
    for (int i = p - 1; i >= 0; i--) cnt[fail[i]] += cnt[i];
  }
  void set(string s, int x) {
    init();
    int len = s.size();
    for (int i = 0; i < len; i++) {
      insert(s[i]);
      dp[x][x + i] = dp[x][x + i - 1] + num[last];
    }
  }
} pam;
void solve(string s) {
  for (int i = 1; i < s.length(); i++) {
    pam.init();
    string ss(s, i, s.length());
    pam.set(ss, i);
  }
}
int main() {
  string s;
  cin >> s;
  solve("#" + s);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    cout << dp[l][r] << endl;
  }
  return 0;
}
