Problem: Ildar took a band (a thin strip of cloth) and colored it. Formally, the band has n cells, each of them is colored into one of 26 colors, so we can denote each color with one of the lowercase letters of English alphabet. 

Ildar decided to take some segment of the band [l, r] (1 ≤ l ≤ r ≤ n) he likes and cut it from the band. So he will create a new band that can be represented as a string t = s_l s_{l+1} … s_r.

After that Ildar will play the following game: he cuts the band t into some new bands and counts the number of different bands among them. Formally, Ildar chooses 1 ≤ k ≤ |t| indexes 1 ≤ i_1 < i_2 < … < i_k = |t| and cuts t to k bands-strings t_1 t_2 … t_{i_1}, t_{i_1 + 1} … t_{i_2}, …, {t_{i_{k-1} + 1}} … t_{i_k} and counts the number of different bands among them. He wants to know the minimal possible number of different bands he can get under the constraint that at least one band repeats at least two times. The result of the game is this number. If it is impossible to cut t in such a way, the result of the game is -1.

Unfortunately Ildar hasn't yet decided which segment he likes, but he has q segments-candidates [l_1, r_1], [l_2, r_2], ..., [l_q, r_q]. Your task is to calculate the result of the game for each of them.

Input

The first line contains one integer n (1 ≤ n ≤ 200 000) — the length of the band Ildar has.

The second line contains a string s consisting of n lowercase English letters — the band Ildar has.

The third line contains a single integer q (1 ≤ q ≤ 200 000) — the number of segments Ildar has chosen as candidates.

Each of the next q lines contains two integer integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) denoting the ends of the i-th segment.

Output

Output q lines, where the i-th of them should contain the result of the game on the segment [l_i, r_i].

Example

Input

9
abcabcdce
7
1 6
4 7
5 9
6 9
1 9
3 6
4 4


Output

1
-1
4
3
2
2
-1

Note

Consider the first example.

If Ildar chooses the segment [1, 6], he cuts a string t = abcabc. If he cuts t into two bands abc and abc, the band abc repeats two times and the number of different tapes is 1. So, the result of this game is 1.

If Ildar chooses the segment [4, 7], he cuts a string t = abcd. It is impossible to cut this band in such a way that there is at least one band repeating at least two times. So, the result of this game is -1.

If Ildar chooses the segment [3, 6], he cuts a string t = cabc. If he cuts t into three bands c, ab and c, the band c repeats two times and the number of different bands is 2. So, the result of this game is 2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool get_int(T &x) {
  char t = getchar();
  bool neg = false;
  x = 0;
  for (; (t > '9' || t < '0') && t != '-' && t != EOF; t = getchar())
    ;
  if (t == '-') neg = true, t = getchar();
  if (t == EOF) return false;
  for (; t <= '9' && t >= '0'; t = getchar()) x = x * 10 + t - '0';
  if (neg) x = -x;
  return true;
}
template <typename T>
void print_int(T x) {
  if (x < 0) putchar('-'), x = -x;
  short a[20] = {}, sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
const int inf = 0x3f3f3f3f;
const long long Linf = 1ll << 61;
const double pi = acos(-1.0);
const int maxn = 200111;
int lg[maxn];
struct arithseq {
  int st, ed, diff;
  arithseq() {}
  arithseq(int a, int b, int c) {
    st = a;
    ed = b;
    diff = c;
  }
  arithseq shift(const int &x) const { return arithseq(st + x, ed + x, diff); }
  arithseq neg() const { return arithseq(-ed, -st, diff); }
  bool have(const int &x) const {
    return (x >= st && x <= ed && (st - x) % diff == 0);
  }
};
int un(const arithseq &a, const arithseq &b) {
  if (a.diff != b.diff) {
    int ret = 0;
    if ((a.ed - a.st) <= (a.diff << 1)) {
      for (int i = a.st; i <= a.ed; i += a.diff)
        if (b.have(i)) ret++;
    } else if ((b.ed - b.st) <= (b.diff << 1)) {
      for (int i = b.st; i <= b.ed; i += b.diff)
        if (a.have(i)) ret++;
    } else {
      puts("baojing");
    }
    return ret;
  } else if ((b.st - a.st) % a.diff)
    return 0;
  else
    return (min(a.ed, b.ed) - max(a.st, b.st)) / a.diff + 1;
}
struct IPM {
  int sa[maxn], rk[maxn], nrk[maxn], c[maxn], n;
  int v[20][maxn];
  int id[20][maxn], mxl;
  vector<int> sq[20 * maxn];
  void construct(char *s) {
    n = strlen(s) + 1;
    int m = 256;
    int *x = rk, *y = nrk;
    for (int i = 0; i < m; i++) c[i] = 0;
    for (int i = 0; i < n; i++) c[x[i] = s[i]]++;
    for (int i = 1; i < m; i++) c[i] += c[i - 1];
    for (int i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    int tot = 0;
    mxl = 0;
    for (int k = 1; k <= n; k <<= 1) {
      for (int i = 0, nxt = 0; i < n; i = nxt) {
        for (; nxt < n && x[sa[i]] == x[sa[nxt]]; nxt++)
          ;
        tot++;
        sq[tot].clear();
        for (int j = i; j < nxt; j++) {
          sq[tot].push_back(sa[j]);
          id[lg[k]][sa[j]] = tot;
        }
        sort(sq[tot].begin(), sq[tot].end());
      }
      mxl = k;
      int p = 0;
      for (int i = n - k; i < n; i++) y[p++] = i;
      for (int i = 0; i < n; i++)
        if (sa[i] >= k) y[p++] = sa[i] - k;
      for (int i = 0; i < m; i++) c[i] = 0;
      for (int i = 0; i < n; i++) c[x[y[i]]]++;
      for (int i = 1; i < m; i++) c[i] += c[i - 1];
      for (int i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
      swap(x, y);
      x[sa[0]] = 0;
      p = 1;
      for (int i = 1; i < n; i++)
        x[sa[i]] =
            (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
                ? p - 1
                : p++;
      if (p >= n) break;
      m = p;
    }
    int k = 0;
    for (int i = 0; i < n; i++) rk[sa[i]] = i;
    for (int i = 0; i < n - 1; i++) {
      if (k) k--;
      int j = sa[rk[i] - 1];
      while (s[i + k] == s[j + k]) k++;
      v[0][rk[i]] = k;
    }
    for (int i = 1; i < 20; i++) {
      for (int j = 2; j < n; j++) {
        int nxt = j + (1 << i - 1);
        if (nxt >= n)
          v[i][j] = v[i - 1][j];
        else
          v[i][j] = min(v[i - 1][j], v[i - 1][nxt]);
      }
    }
  }
  int qlcp(int l, int r) {
    if (l == r) return n - l;
    l = rk[l];
    r = rk[r];
    if (l > r) swap(l, r);
    l++;
    int t = lg[r - l + 1];
    return min(v[t][l], v[t][r - (1 << t) + 1]);
  }
  int pred(int l, int k, int pos) {
    if (k > mxl) {
      if (pos >= l)
        return l;
      else
        return -1;
    }
    vector<int> &v = sq[id[lg[k]][l]];
    int id = upper_bound(v.begin(), v.end(), pos) - v.begin();
    if (id)
      return v[id - 1];
    else
      return -1;
  }
  int succ(int l, int k, int pos) {
    if (k > mxl) {
      if (pos <= l)
        return l;
      else
        return -1;
    }
    vector<int> &v = sq[id[lg[k]][l]];
    int id = lower_bound(v.begin(), v.end(), pos) - v.begin();
    if (id < (int)v.size())
      return v[id];
    else
      return -1;
  }
  int get_border(int l, int r) {
    int len = r - l + 1, l1, l2, l3, r1, r2, r3, ret = 0;
    for (int k = 1; k < len; k <<= 1) {
      int tl = l, tr = r - k + 1;
      l1 = succ(tr, k, tl);
      if (l1 == -1 || l1 >= l + k || l1 >= r) continue;
      r1 = pred(tl, k, tr);
      if (r1 == -1 || r1 <= tr - k || r1 <= l) continue;
      l2 = succ(tr, k, l1 + 1);
      l3 = pred(tr, k, l + k - 1);
      r2 = pred(tl, k, r1 - 1);
      r3 = succ(tl, k, max(l + 1, tr - k + 1));
      arithseq k1, k2;
      if (l2 != -1)
        k1 = arithseq(l1 - l, l3 - l, l2 - l1);
      else
        k1 = arithseq(l1 - l, l1 - l, 1);
      if (r2 != -1)
        k2 = arithseq(tr - r1, tr - r3, r1 - r2);
      else
        k2 = arithseq(tr - r1, tr - r1, 1);
      ret += un(k1, k2);
    }
    return ret;
  }
} thor;
struct suffix {
  int sa[maxn], rk[maxn], nrk[maxn], c[maxn], n;
  int v[20][maxn];
  void construct(char *s) {
    n = strlen(s) + 1;
    int m = 256;
    int *x = rk, *y = nrk;
    for (int i = 0; i < m; i++) c[i] = 0;
    for (int i = 0; i < n; i++) c[x[i] = s[i]]++;
    for (int i = 1; i < m; i++) c[i] += c[i - 1];
    for (int i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for (int k = 1; k <= n; k <<= 1) {
      int p = 0;
      for (int i = n - k; i < n; i++) y[p++] = i;
      for (int i = 0; i < n; i++)
        if (sa[i] >= k) y[p++] = sa[i] - k;
      for (int i = 0; i < m; i++) c[i] = 0;
      for (int i = 0; i < n; i++) c[x[y[i]]]++;
      for (int i = 1; i < m; i++) c[i] += c[i - 1];
      for (int i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
      swap(x, y);
      x[sa[0]] = 0;
      p = 1;
      for (int i = 1; i < n; i++)
        x[sa[i]] =
            (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
                ? p - 1
                : p++;
      if (p >= n) break;
      m = p;
    }
    int k = 0;
    for (int i = 0; i < n; i++) rk[sa[i]] = i;
    for (int i = 0; i < n - 1; i++) {
      if (k) k--;
      int j = sa[rk[i] - 1];
      while (s[i + k] == s[j + k]) k++;
      v[0][rk[i]] = k;
    }
    for (int i = 1; i < 20; i++) {
      for (int j = 2; j < n; j++) {
        int nxt = j + (1 << i - 1);
        if (nxt >= n)
          v[i][j] = v[i - 1][j];
        else
          v[i][j] = min(v[i - 1][j], v[i - 1][nxt]);
      }
    }
  }
  int qlcp(int l, int r) {
    if (l == r) return n - l;
    l = rk[l];
    r = rk[r];
    if (l > r) swap(l, r);
    l++;
    int t = lg[r - l + 1];
    return min(v[t][l], v[t][r - (1 << t) + 1]);
  }
} loki;
char s[maxn];
int n, q, sum[maxn][26], v[20][maxn];
struct solver {
  int f[maxn], a[maxn];
  void prework(int coef) {
    for (int i = 1; i <= n; i++) {
      f[i] = i;
      a[i] = (coef == 1 ? inf : -inf);
    }
  }
  int gf(int x) { return x == f[x] ? x : f[x] = gf(f[x]); }
  void modify(int l, int r, int v) {
    while (true) {
      l = gf(l);
      if (l && l <= r) {
        a[l] = l + v;
        f[l] = l + 1;
      } else
        break;
    }
  }
  void construct() {
    for (int i = 1; i <= n; i++) v[0][i] = a[i];
    for (int i = 1; i < 20; i++) {
      for (int j = 1; j <= n; j++) {
        int nxt = j + (1 << i - 1);
        if (nxt <= n)
          v[i][j] = min(v[i - 1][j], v[i - 1][nxt]);
        else
          v[i][j] = v[i - 1][j];
      }
    }
  }
};
solver mnl, mnr;
int query(int l, int r) {
  int t = lg[r - l + 1];
  return min(v[t][l], v[t][r - (1 << t) + 1]);
}
int lcp(int a, int b) { return thor.qlcp(a - 1, b - 1); }
int lcs(int a, int b) { return loki.qlcp(n - a, n - b); }
vector<int> pr[maxn];
bool per(int l, int r) {
  int len = r - l + 1;
  for (int i = 0; i < (int)pr[len].size(); i++) {
    if (lcp(l, l + len / pr[len][i]) + len / pr[len][i] >= len) return true;
  }
  return false;
}
int main() {
  for (int i = 2; i < maxn; i++) {
    lg[i] = lg[i >> 1] + 1;
    if (pr[i].size() == 0) {
      for (int j = i; j < maxn; j += i) pr[j].push_back(i);
    }
  }
  scanf("%d%s", &n, s + 1);
  thor.construct(s + 1);
  reverse(s + 1, s + n + 1);
  loki.construct(s + 1);
  reverse(s + 1, s + n + 1);
  for (int i = 1; i <= n; i++) {
    memcpy(sum[i], sum[i - 1], sizeof(sum[i]));
    sum[i][s[i] - 'a']++;
  }
  mnl.prework(1);
  mnr.prework(-1);
  for (int len = 1; len + len <= n; len++) {
    for (int i = len; i + len <= n; i += len) {
      int j = i + len, l1 = lcs(i, j), l2 = lcp(i + 1, j + 1);
      if (l1 + l2 < len || l1 == 0) continue;
      int l = i - min(len, l1) + 1, r = i - max(1, len - l2) + 1;
      mnl.modify(l, r, len + len - 1);
      mnr.modify(l + len + len - 1, r + len + len - 1, -len - len + 1);
    }
  }
  mnl.construct();
  get_int(q);
  for (int i = 1, l, r; i <= q; i++) {
    (get_int(l) && get_int(r));
    int ans = -1;
    for (int j = 0; j < 26 && ans == -1; j++)
      if (sum[r][j] - sum[l - 1][j] > 1) ans = 0;
    if (ans == -1)
      puts("-1");
    else if (per(l, r))
      puts("1");
    else if (mnl.a[l] < r || mnr.a[r] > l || thor.get_border(l - 1, r - 1))
      puts("2");
    else if (sum[r][s[l] - 'a'] > sum[l][s[l] - 'a'] ||
             sum[r - 1][s[r] - 'a'] > sum[l - 1][s[r] - 'a'] ||
             query(l, r) <= r)
      puts("3");
    else
      puts("4");
  }
  return 0;
}