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 = 5010;
char str[maxn];
int p[maxn][maxn];
int dp[maxn][maxn];
int dfs2(int x, int y) {
  if (p[x][y] != -1) return p[x][y];
  if (x == y) return p[x][y] = 1;
  if (x + 1 == y)
    p[x][y] = str[x] == str[y];
  else
    p[x][y] = (str[x] == str[y]) && dfs2(x + 1, y - 1);
  return p[x][y];
}
int dfs(int x, int y) {
  if (x > y) return 0;
  if (dp[x][y] != -1) return dp[x][y];
  if (x == y) {
    return dp[x][y] = 1;
  }
  dp[x][y] = dfs(x + 1, y) + dfs(x, y - 1) - dfs(x + 1, y - 1);
  dp[x][y] += dfs2(x, y);
  return dp[x][y];
}
int main() {
  scanf("%s", str);
  int len = strlen(str);
  memset(dp, -1, sizeof(dp));
  memset(p, -1, sizeof(p));
  dfs(0, len - 1);
  int T;
  cin >> T;
  while (T--) {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%d\n", dfs(x - 1, y - 1));
  }
  return 0;
}
