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 dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long int lxt() {
  long long int aaa;
  scanf("%I64d", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
long long int dp[5010][5010];
const long long MOD[] = {1000000007LL, 2117566807LL, 1000000007LL,
                         1000000037LL};
const long long BASE[] = {1572872831LL, 1971536491LL, 1000003LL, 31};
class stringhash {
 public:
  long long int base, mod, len, ar[5010], P[5010];
  void GenerateFH(string s, long long int b, long long int m) {
    base = b;
    mod = m;
    len = s.length();
    P[0] = 1;
    long long h = 0;
    for (int i = 1; i <= len; i++)
      P[i] = ((long long int)P[i - 1] * (long long int)base) % mod;
    for (int i = 0; i < len; i++) {
      h = (h * base) + s[i];
      h %= mod;
      ar[i] = h;
    }
  }
  int Fhash(int x, int y) {
    long long int h = ar[y];
    if (x > 0) {
      h = (h - (long long int)P[y - x + 1] * (long long int)ar[x - 1]) % mod;
      if (h < 0) h += mod;
    }
    return h;
  }
  void GenerateRH(string s, long long int b, long long int m) {
    base = b;
    mod = m;
    len = s.length();
    P[0] = 1;
    long long int h = 0;
    for (int i = 1; i <= len; i++)
      P[i] = ((long long int)P[i - 1] * (long long int)base) % mod;
    for (int i = len - 1; i >= 0; i--) {
      h = (h * base) + s[i];
      h %= mod;
      ar[i] = h;
    }
  }
  long long int Rhash(int x, int y) {
    long long h = ar[x];
    if (y < len - 1) {
      h = (h - (long long int)P[y - x + 1] * (long long int)ar[y + 1]) % mod;
      if (h < 0) h += mod;
    }
    return h;
  }
};
stringhash s1, s2;
int main() {
  string s;
  cin >> s;
  s = '&' + s;
  s1.GenerateFH(s, BASE[0], MOD[0]);
  s2.GenerateRH(s, BASE[0], MOD[0]);
  int n = s.length();
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      dp[i][j] = (s1.Fhash(i, j) == s2.Rhash(i, j));
      dp[i][j] = dp[i][j] + dp[i][j - 1];
    }
  }
  for (int i = 1; i < n; i++)
    for (int j = 1; j < n; j++) dp[i][j] += dp[i - 1][j];
  int q = nxt();
  while (q--) {
    int a = nxt();
    int b = nxt();
    int ans = dp[b][b] + dp[a - 1][a - 1] - dp[a - 1][b] - dp[b][a - 1];
    printf("%d\n", ans);
  }
  return 0;
}
