#include <bits/stdc++.h>
using namespace std;
const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
  if (rsz <= 0) {
    ptr = 0;
    rsz = fread(bf, 1, bfsz, stdin);
    if (rsz <= 0) return EOF;
  }
  --rsz;
  return bf[ptr++];
}
void ga(char &c) {
  c = EOF;
  while (!isalpha(c)) c = gc();
}
int gs(char s[]) {
  int l = 0;
  char c = gc();
  while (isspace(c)) c = gc();
  while (c != EOF && !isspace(c)) {
    s[l++] = c;
    c = gc();
  }
  s[l] = '\0';
  return l;
}
template <class T>
bool gi(T &v) {
  v = 0;
  char c = gc();
  while (c != EOF && c != '-' && !isdigit(c)) c = gc();
  if (c == EOF) return false;
  bool neg = c == '-';
  if (neg) c = gc();
  while (isdigit(c)) {
    v = v * 10 + c - '0';
    c = gc();
  }
  if (neg) v = -v;
  return true;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int toInt(char xx) { return xx - '0'; }
char toChar(int xx) { return xx + '0'; }
int isDigit(char xx) { return ('0' <= xx && xx <= '9'); }
void swapInt(int &x, int &y) {
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
}
class bigint {
 public:
  int a[100];
  void add(bigint &y);
  void print();
};
void bigint::add(bigint &y) {
  int i, d;
  if (a[0] < y.a[0]) a[0] = y.a[0];
  d = 0;
  for (i = 1; i <= a[0]; i++) {
    a[i] += y.a[i] + d;
    d = a[i] / 1000000;
    a[i] %= 1000000;
  }
  while (d > 0) {
    a[0]++;
    a[a[0]] = d % 1000000;
    d /= 1000000;
  }
}
void bigint::print() {
  int i, n = a[0];
  cout << a[n];
  for (i = n - 1; i >= 1; i--) {
    if (a[i] < 10)
      cout << "00000";
    else if (a[i] < 100)
      cout << "0000";
    else if (a[i] < 1000)
      cout << "000";
    else if (a[i] < 10000)
      cout << "00";
    else if (a[i] < 100000)
      cout << "0";
    cout << a[i];
  }
}
int exitInput = 0;
int ntest = 1, itest = 1;
void read();
void solve();
int main() {
  clock_t t1, t2;
  t1 = clock();
  for (itest = 1; itest <= ntest; ++itest) {
    read();
    if (exitInput) {
      break;
    }
    if (itest > 1) {
    }
    solve();
  }
  t2 = clock();
  return 0;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const long long Mod = 1e9 + 7;
const int maxn = 3 * 1e5 + 2;
int n, a[maxn], idx[maxn];
void dfs(int u, int end, int &res) {
  if (u == end) {
    return;
  }
  int v = u + 1;
  if (idx[v] < idx[u]) {
    res++;
  }
  dfs(v, end, res);
}
void read() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    idx[a[i]] = i;
  }
}
void solve() {
  int Q, id, x, y, res;
  cin >> Q;
  while (Q--) {
    cin >> id >> x >> y;
    if (id == 2) {
      swap(a[x], a[y]);
      idx[a[x]] = x;
      idx[a[y]] = y;
    } else if (id == 1) {
      res = 1;
      dfs(x, y, res);
      cout << res << endl;
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int bfsz = 1 << 16;
char bf[bfsz + 5];
int rsz = 0;
int ptr = 0;
char gc() {
  if (rsz <= 0) {
    ptr = 0;
    rsz = fread(bf, 1, bfsz, stdin);
    if (rsz <= 0) return EOF;
  }
  --rsz;
  return bf[ptr++];
}
void ga(char &c) {
  c = EOF;
  while (!isalpha(c)) c = gc();
}
int gs(char s[]) {
  int l = 0;
  char c = gc();
  while (isspace(c)) c = gc();
  while (c != EOF && !isspace(c)) {
    s[l++] = c;
    c = gc();
  }
  s[l] = '\0';
  return l;
}
template <class T>
bool gi(T &v) {
  v = 0;
  char c = gc();
  while (c != EOF && c != '-' && !isdigit(c)) c = gc();
  if (c == EOF) return false;
  bool neg = c == '-';
  if (neg) c = gc();
  while (isdigit(c)) {
    v = v * 10 + c - '0';
    c = gc();
  }
  if (neg) v = -v;
  return true;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
int toInt(char xx) { return xx - '0'; }
char toChar(int xx) { return xx + '0'; }
int isDigit(char xx) { return ('0' <= xx && xx <= '9'); }
void swapInt(int &x, int &y) {
  x = x ^ y;
  y = x ^ y;
  x = x ^ y;
}
class bigint {
 public:
  int a[100];
  void add(bigint &y);
  void print();
};
void bigint::add(bigint &y) {
  int i, d;
  if (a[0] < y.a[0]) a[0] = y.a[0];
  d = 0;
  for (i = 1; i <= a[0]; i++) {
    a[i] += y.a[i] + d;
    d = a[i] / 1000000;
    a[i] %= 1000000;
  }
  while (d > 0) {
    a[0]++;
    a[a[0]] = d % 1000000;
    d /= 1000000;
  }
}
void bigint::print() {
  int i, n = a[0];
  cout << a[n];
  for (i = n - 1; i >= 1; i--) {
    if (a[i] < 10)
      cout << "00000";
    else if (a[i] < 100)
      cout << "0000";
    else if (a[i] < 1000)
      cout << "000";
    else if (a[i] < 10000)
      cout << "00";
    else if (a[i] < 100000)
      cout << "0";
    cout << a[i];
  }
}
int exitInput = 0;
int ntest = 1, itest = 1;
void read();
void solve();
int main() {
  clock_t t1, t2;
  t1 = clock();
  for (itest = 1; itest <= ntest; ++itest) {
    read();
    if (exitInput) {
      break;
    }
    if (itest > 1) {
    }
    solve();
  }
  t2 = clock();
  return 0;
}
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const long long Mod = 1e9 + 7;
const int maxn = 3 * 1e5 + 2;
int n, a[maxn], idx[maxn];
void dfs(int u, int end, int &res) {
  if (u == end) {
    return;
  }
  int v = u + 1;
  if (idx[v] < idx[u]) {
    res++;
  }
  dfs(v, end, res);
}
void read() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    idx[a[i]] = i;
  }
}
void solve() {
  int Q, id, x, y, res;
  cin >> Q;
  while (Q--) {
    cin >> id >> x >> y;
    if (id == 2) {
      swap(a[x], a[y]);
      idx[a[x]] = x;
      idx[a[y]] = y;
    } else if (id == 1) {
      res = 1;
      dfs(x, y, res);
      cout << res << endl;
    }
  }
}

```