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;
const int maxn = 3e5 + 10;
int n, q;
int a[maxn];
int pos[maxn];
int cur;
int seg[4 * maxn];
void update(int idx, int i, int j, int pos, int val) {
  if (i == j) {
    seg[idx] += val;
    return;
  }
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  if (pos <= mid)
    update(left, i, mid, pos, val);
  else
    update(right, mid + 1, j, pos, val);
  seg[idx] = seg[left] + seg[right];
}
int query(int idx, int i, int j, int l, int r) {
  if (l > r || i > j || j < l || r < i) return 0;
  if (l <= i && j <= r) return seg[idx];
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  return query(left, i, mid, l, r) + query(right, mid + 1, j, l, r);
}
struct comp {
  int L, R;
} C[maxn * 10];
struct node {
  int id, lazy;
} tree[4 * maxn];
void refresh(int idx, int i, int j) {
  if (!tree[idx].lazy) return;
  int left = idx << 1;
  int right = left | 1;
  if (i != j) {
    tree[left].lazy = tree[idx].lazy;
    tree[right].lazy = tree[idx].lazy;
  } else
    tree[idx].id = tree[idx].lazy;
  tree[idx].lazy = 0;
}
void change(int idx, int i, int j, int l, int r, int id) {
  refresh(idx, i, j);
  if (l > r || i > j || j < l || r < i) return;
  if (l <= i && j <= r) {
    tree[idx].lazy = id;
    refresh(idx, i, j);
    return;
  }
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  change(left, i, mid, l, r, id);
  change(right, mid + 1, j, l, r, id);
}
int getId(int idx, int i, int j, int pos) {
  refresh(idx, i, j);
  if (i == j) return tree[idx].id;
  int mid = (i + j) >> 1;
  int left = idx << 1;
  int right = left | 1;
  if (pos <= mid) return getId(left, i, mid, pos);
  return getId(right, mid + 1, j, pos);
}
void build() {
  for (int i = 1; i <= n; ++i) pos[a[i]] = i;
  for (int i = 1; i <= n; ++i) {
    int L = i;
    int j = i;
    ++cur;
    while (j + 1 <= n && pos[j] < pos[j + 1]) ++j;
    int R = j;
    i = j;
    C[cur] = {L, R};
    update(1, 1, n, L, +1);
    change(1, 1, n, L, R, cur);
  }
}
inline int getAns(int x, int y) {
  int X = getId(1, 1, n, x);
  int Y = getId(1, 1, n, y);
  if (X == Y) return 1;
  return 2 + query(1, 1, n, C[X].R + 1, C[Y].L - 1);
}
inline void Break(int mid) {
  int ID = getId(1, 1, n, mid);
  int L = C[ID].L;
  int R = C[ID].R;
  assert(query(1, 1, n, L, L) == 1);
  update(1, 1, n, L, -1);
  if (L <= mid - 1) {
    ++cur;
    C[cur] = {L, mid - 1};
    update(1, 1, n, L, +1);
    change(1, 1, n, L, mid - 1, cur);
  }
  if (mid + 1 <= R) {
    ++cur;
    C[cur] = {mid + 1, R};
    update(1, 1, n, mid + 1, +1);
    change(1, 1, n, mid + 1, R, cur);
  }
  C[ID] = {mid, mid};
  update(1, 1, n, mid, +1);
}
inline void complete(int mid) {
  int Id = getId(1, 1, n, mid);
  int L = C[Id].L;
  int R = C[Id].R;
  update(1, 1, n, L, -1);
  ++cur;
  if (mid + 1 <= n && pos[mid] < pos[mid + 1]) {
    int x = getId(1, 1, n, mid + 1);
    if (x != Id) {
      update(1, 1, n, C[x].L, -1);
      R = C[x].R;
    }
  }
  if (mid - 1 >= 1 && pos[mid - 1] < pos[mid]) {
    int x = getId(1, 1, n, mid - 1);
    if (x != Id) {
      update(1, 1, n, C[x].L, -1);
      L = C[x].L;
    }
  }
  C[cur] = {L, R};
  update(1, 1, n, L, +1);
  change(1, 1, n, L, R, cur);
}
void upd(int x, int y) {
  Break(x);
  Break(y);
  swap(pos[x], pos[y]);
  complete(x);
  complete(y);
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build();
  cin >> q;
  while (q--) {
    int type, x, y;
    cin >> type >> x >> y;
    if (type == 1)
      cout << getAns(x, y) << '\n';
    else {
      upd(a[x], a[y]);
      swap(a[x], a[y]);
    }
  }
  return 0;
}
