The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!

There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a1, a2, ..., an of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i1 < i2 < ... < ik, that ai1 = x, ai2 = x + 1, ..., aik - 1 = y - 1, aik = y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1, 2, 3, ..., n.

If we can't shave beavers from x to y in one session, then we can split these beavers into groups [x, p1], [p1 + 1, p2], ..., [pm + 1, y] (x ≤ p1 < p2 < ... < pm < y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m + 1 working sessions to shave beavers from x to y.

All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types: 

  * what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive? 
  * two beavers on positions x and y (the beavers ax and ay) swapped. 



You can assume that any beaver can be shaved any number of times.

Input

The first line contains integer n — the total number of beavers, 2 ≤ n. The second line contains n space-separated integers — the initial beaver permutation.

The third line contains integer q — the number of queries, 1 ≤ q ≤ 105. The next q lines contain the queries. Each query i looks as pi xi yi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1 ≤ xi < yi ≤ n.

  * to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem B1); 
  * to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems B1+B2). 



Note that the number of queries q is limited 1 ≤ q ≤ 105 in both subproblem B1 and subproblem B2.

Output

For each query with pi = 1, print the minimum number of Beavershave 5000 sessions.

Examples

Input

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5


Output

2
1
3
5
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / __gcd(a, b));
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
template <class T>
string itoa(T a) {
  if (!a) return "0";
  string ret;
  for (T i = a; i > 0; i = i / 10) ret.push_back((i % 10) + 48);
  reverse(ret.begin(), ret.end());
  return ret;
}
double log(double N, double B) { return (log10l(N)) / (log10l(B)); }
vector<string> token(string a, string b) {
  const char *q = a.c_str();
  while (count(b.begin(), b.end(), *q)) q++;
  vector<string> oot;
  while (*q) {
    const char *e = q;
    while (*e && !count(b.begin(), b.end(), *e)) e++;
    oot.push_back(string(q, e));
    q = e;
    while (count(b.begin(), b.end(), *q)) q++;
  }
  return oot;
}
string toBin(int n) {
  string s;
  for (int i = 10 - 1; i >= 0; i--) s += ((bool)(n & (1 << i)) + '0');
  return s;
}
int tree[2048561];
int arr[300002], p[300002];
void init(int node, int b, int e) {
  if (b == e) {
    tree[node] = 0;
    return;
  }
  int Left = node * 2;
  int Right = node * 2 + 1;
  int mid = (b + e) / 2;
  init(Left, b, mid);
  init(Right, mid + 1, e);
  tree[node] = tree[Left] + tree[Right] + (p[mid + 1] < p[mid]);
}
struct trio {
  int val, b, e;
  trio(int val, int b, int e) : val(val), b(b), e(e) {}
  trio() {}
};
trio query(int node, int b, int e, int i, int j) {
  if (i > e || j < b) return trio(0, -1, -1);
  if (b >= i && e <= j) {
    return trio(tree[node], b, e);
  }
  int Left = node * 2;
  int Right = node * 2 + 1;
  int mid = (b + e) / 2;
  trio PL = query(Left, b, mid, i, j);
  trio PR = query(Right, mid + 1, e, i, j);
  if (PL.b == -1) return PR;
  if (PR.b == -1) return PL;
  int val = PL.val + PR.val + (p[PR.b] < p[PL.e]);
  return trio(val, b, e);
}
void update(int node, int b, int e, int idx) {
  if (idx > e || idx < b) return;
  if (b == e) return;
  int Left = node * 2;
  int Right = node * 2 + 1;
  int mid = (b + e) / 2;
  update(Left, b, mid, idx);
  update(Right, mid + 1, e, idx);
  tree[node] = tree[Left] + tree[Right] + (p[mid + 1] < p[mid]);
}
int n;
int query(int b, int e) { return query(1, 1, n, b, e).val; }
int main() {
  cin >> n;
  for (int i = 1; i <= (int)n; i++) {
    scanf("%d", &arr[i]);
    p[arr[i]] = i;
  }
  init(1, 1, n);
  int q;
  cin >> q;
  while (q--) {
    int c, a, b;
    scanf("%d", &c);
    scanf("%d%d", &a, &b);
    if (c == 1) {
      int ret = query(a, b) + 1;
      printf("%d\n", ret);
      ;
    } else {
      swap(arr[a], arr[b]);
      int p1 = p[arr[a]];
      int p2 = p[arr[b]];
      p[arr[a]] = p2;
      update(1, 1, n, arr[a]);
      p[arr[b]] = p1;
      update(1, 1, n, arr[b]);
    }
  }
  return 0;
}
