Recently Lynyrd and Skynyrd went to a shop where Lynyrd bought a permutation p of length n, and Skynyrd bought an array a of length m, consisting of integers from 1 to n. 

Lynyrd and Skynyrd became bored, so they asked you q queries, each of which has the following form: "does the subsegment of a from the l-th to the r-th positions, inclusive, have a subsequence that is a cyclic shift of p?" Please answer the queries.

A permutation of length n is a sequence of n integers such that each integer from 1 to n appears exactly once in it.

A cyclic shift of a permutation (p_1, p_2, …, p_n) is a permutation (p_i, p_{i + 1}, …, p_{n}, p_1, p_2, …, p_{i - 1}) for some i from 1 to n. For example, a permutation (2, 1, 3) has three distinct cyclic shifts: (2, 1, 3), (1, 3, 2), (3, 2, 1).

A subsequence of a subsegment of array a from the l-th to the r-th positions, inclusive, is a sequence a_{i_1}, a_{i_2}, …, a_{i_k} for some i_1, i_2, …, i_k such that l ≤ i_1 < i_2 < … < i_k ≤ r.

Input

The first line contains three integers n, m, q (1 ≤ n, m, q ≤ 2 ⋅ 10^5) — the length of the permutation p, the length of the array a and the number of queries.

The next line contains n integers from 1 to n, where the i-th of them is the i-th element of the permutation. Each integer from 1 to n appears exactly once.

The next line contains m integers from 1 to n, the i-th of them is the i-th element of the array a.

The next q lines describe queries. The i-th of these lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m), meaning that the i-th query is about the subsegment of the array from the l_i-th to the r_i-th positions, inclusive.

Output

Print a single string of length q, consisting of 0 and 1, the digit on the i-th positions should be 1, if the subsegment of array a from the l_i-th to the r_i-th positions, inclusive, contains a subsequence that is a cyclic shift of p, and 0 otherwise.

Examples

Input

3 6 3
2 1 3
1 2 3 1 2 3
1 5
2 6
3 5


Output

110


Input

2 4 3
2 1
1 1 2 2
1 2
2 3
3 4


Output

010

Note

In the first example the segment from the 1-st to the 5-th positions is 1, 2, 3, 1, 2. There is a subsequence 1, 3, 2 that is a cyclic shift of the permutation. The subsegment from the 2-nd to the 6-th positions also contains a subsequence 2, 1, 3 that is equal to the permutation. The subsegment from the 3-rd to the 5-th positions is 3, 1, 2, there is only one subsequence of length 3 (3, 1, 2), but it is not a cyclic shift of the permutation.

In the second example the possible cyclic shifts are 1, 2 and 2, 1. The subsegment from the 1-st to the 2-nd positions is 1, 1, its subsequences are not cyclic shifts of the permutation. The subsegment from the 2-nd to the 3-rd positions is 1, 2, it coincides with the permutation. The subsegment from the 3 to the 4 positions is 2, 2, its subsequences are not cyclic shifts of the permutation.
#include <bits/stdc++.h>
using namespace std;
inline int readChar();
template <class T = int>
inline T readInt();
template <class T>
inline void writeInt(T x, char end = 0);
inline void writeChar(int x);
inline void writeWord(const char* s);
static const int buf_size = 4096;
inline int getChar() {
  static char buf[buf_size];
  static int len = 0, pos = 0;
  if (pos == len) pos = 0, len = fread(buf, 1, buf_size, stdin);
  if (pos == len) return -1;
  return buf[pos++];
}
inline int readChar() {
  int c = getChar();
  while (c <= 32) c = getChar();
  return c;
}
template <class T>
inline T readInt() {
  int s = 1, c = readChar();
  T x = 0;
  if (c == '-') s = -1, c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return s == 1 ? x : -x;
}
static int write_pos = 0;
static char write_buf[buf_size];
inline void writeChar(int x) {
  if (write_pos == buf_size)
    fwrite(write_buf, 1, buf_size, stdout), write_pos = 0;
  write_buf[write_pos++] = x;
}
template <class T>
inline void writeInt(T x, char end) {
  if (x < 0) writeChar('-'), x = -x;
  char s[24];
  int n = 0;
  while (x || !n) s[n++] = '0' + x % 10, x /= 10;
  while (n--) writeChar(s[n]);
  if (end) writeChar(end);
}
inline void writeWord(const char* s) {
  while (*s) writeChar(*s++);
}
struct Flusher {
  ~Flusher() {
    if (write_pos) fwrite(write_buf, 1, write_pos, stdout), write_pos = 0;
  }
} flusher;
inline void readWord(char* s) {
  int c = readChar();
  while (c > 32) *s++ = c, c = getChar();
  *s = 0;
}
inline void writeDouble(double x, int output_len) {
  if (x < 0) writeChar('-'), x = -x;
  int t = (int)x;
  writeInt(t), x -= t;
  writeChar('.');
  for (int i = output_len - 1; i > 0; i--) {
    x *= 10;
    t = min(9, (int)x);
    writeChar('0' + t), x -= t;
  }
  x *= 10;
  t = min(9, (int)(x + 0.5));
  writeChar('0' + t);
}
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using ull = unsigned long long;
using pnt = complex<ld>;
istream& operator>>(istream& in, pnt& p) {
  ld a, b;
  in >> a >> b;
  p = {a, b};
  return in;
}
ld crs(pnt a, pnt b) { return a.real() * b.imag() - a.imag() * b.real(); }
ld dot(pnt a, pnt b) { return a.real() * b.real() + a.imag() * b.imag(); }
const int M = 998244353;
const int mod = (int)1e9 + 7;
const int inf = (int)1e9 + 100;
const ll inf64 = (ll)1e18L + 100;
const int di[] = {0, 1, -1, 0};
const int dj[] = {1, 0, 0, -1};
const int LG = 20;
const int nax = (int)5e5 + 100;
int n, m, q;
int p[nax];
int a[nax];
int inp[nax];
int up[LG][nax];
int fst[nax];
int kek[nax];
struct segTree {
  int n;
  vi t;
  segTree(int n) : n(n), t(2 * n, inf) {}
  void upd(int v, int x) {
    t[v += n] = x;
    for (v >>= 1; v; v >>= 1) {
      t[v] = min(t[v * 2], t[v * 2 + 1]);
    }
  }
  int get(int l, int r) {
    int ret = inf;
    for (l += n, r += n; l <= r; l = (l + 1) >> 1, r = (r - 1) >> 1) {
      if (l & 1) {
        ret = min(ret, t[l]);
      }
      if (!(r & 1)) {
        ret = min(ret, t[r]);
      }
    }
    return ret;
  }
};
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
    inp[p[i]] = i;
  }
  p[n] = p[0];
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < LG; i++) {
    up[i][m] = m;
  }
  fill_n(fst, n, m);
  for (int i = m - 1; i >= 0; i--) {
    up[0][i] = fst[p[inp[a[i]] + 1]];
    fst[a[i]] = i;
  }
  for (int j = 1; j < LG; j++) {
    for (int i = 0; i < m; i++) {
      up[j][i] = up[j - 1][up[j - 1][i]];
    }
  }
  for (int i = 0; i < m; i++) {
    int v = i;
    int rem = n - 1;
    for (int j = LG - 1; j >= 0; j--) {
      if ((1 << j) <= rem) {
        rem -= (1 << j);
        v = up[j][v];
      }
    }
    kek[i] = v;
  }
  segTree t(m);
  for (int i = 0; i < m; i++) {
    t.upd(i, kek[i]);
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << (t.get(l, r) <= r);
  }
}
