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 = 5005;
int n, m;
char ch[maxn];
int a[maxn];
struct Node {
  int l, r;
  int id;
  bool operator<(const Node& t) const { return r < t.r; }
} p[1000 * 1000];
vector<int> edge[maxn];
int ans[maxn * 1000];
int tree[maxn << 2];
int LL(int x) { return x << 1; }
int RR(int x) { return x << 1 | 1; }
inline void push_down(int k) {
  if (tree[k]) {
    tree[LL(k)] += tree[k];
    tree[RR(k)] += tree[k];
    tree[k] = 0;
  }
}
inline void update(int L, int R, int l, int r, int k) {
  if (l <= L && R <= r) {
    tree[k]++;
    return;
  }
  push_down(k);
  int mid = (L + R) >> 1;
  if (l <= mid) update(L, mid, l, r, LL(k));
  if (r > mid) update(mid + 1, R, l, r, RR(k));
}
inline int query(int L, int R, int id, int k) {
  if (L == id && R == id) {
    return tree[k];
  }
  push_down(k);
  int mid = (L + R) >> 1;
  if (id <= mid)
    return query(L, mid, id, LL(k));
  else
    return query(mid + 1, R, id, RR(k));
}
int main() {
  int i, j, k;
  while (scanf("%s", ch) != EOF) {
    m = strlen(ch);
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%d%d", &p[i].l, &p[i].r);
      p[i].l--;
      p[i].r--;
      p[i].id = i;
    }
    sort(p, p + n);
    for (i = 0; i < m; i++) {
      for (j = i, k = i; j < m && k >= 0; j++, k--) {
        if (ch[j] == ch[k]) {
          edge[j].push_back(k);
        } else {
          break;
        }
      }
      for (j = i, k = i - 1; j < m && k >= 0; j++, k--) {
        if (ch[j] == ch[k]) {
          edge[j].push_back(k);
        } else {
          break;
        }
      }
    }
    for (i = 0, k = 0; i < m; i++) {
      for (j = 0; j < edge[i].size(); j++) {
        update(0, m, 0, edge[i][j], 1);
      }
      while (k < n && p[k].r <= i) {
        ans[p[k].id] = query(0, m, p[k].l, 1);
        k++;
      }
    }
    for (i = 0; i < n; i++) {
      printf("%d\n", ans[i]);
    }
  }
  return 0;
}
