#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[8][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a > b ? a : b;
  if (c != -1234567) ans = max1(ans, c);
  if (d != -1234567) ans = max1(ans, d);
  return ans;
}
template <class T>
inline T min1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a < b ? a : b;
  if (c != -1234567) ans = min(ans, c);
  if (d != -1234567) ans = min(ans, d);
  return ans;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
template <class T>
inline T lb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / abs1(a) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
inline int bitnum(unsigned long long nValue) {
  nValue = ((0xaaaaaaaaaaaaaaaaull & nValue) >> 1) +
           (0x5555555555555555ull & nValue);
  nValue = ((0xccccccccccccccccull & nValue) >> 2) +
           (0x3333333333333333ull & nValue);
  nValue = ((0xf0f0f0f0f0f0f0f0ull & nValue) >> 4) +
           (0x0f0f0f0f0f0f0f0full & nValue);
  nValue = ((0xff00ff00ff00ff00ull & nValue) >> 8) +
           (0x00ff00ff00ff00ffull & nValue);
  nValue = ((0xffff0000ffff0000ull & nValue) >> 16) +
           (0x0000ffff0000ffffull & nValue);
  nValue = ((0xffffffff00000000ull & nValue) >> 32) +
           (0x00000000ffffffffull & nValue);
  return nValue;
}
long long pow(long long n, long long m, long long mod = 0) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int maxn = 500100;
template <class t>
struct tree_array {
  t num[maxn], n, bigwei;
  tree_array(int nn = maxn) {
    n = nn;
    while (lb(nn) != nn) nn -= lb(nn);
    bigwei = nn;
  }
  void upd(int no, t add) {
    while (no <= n) {
      num[no] += add;
      no += lb(no);
    }
  }
  t que(int no) {
    t ans = 0;
    while (no) {
      ans += num[no];
      no -= lb(no);
    }
    return ans;
  }
  int getrank(t x) {
    int ans = 0, ranwei = bigwei;
    t rank = 0;
    while (ranwei) {
      if (rank + num[ranwei + ans] < x) {
        ans += ranwei;
        rank += num[ans];
      }
      ranwei >>= 1;
    }
    return ans + 1;
  }
};
tree_array<int> ta;
int n, orig[maxn];
int nq;
int trans[maxn];
inline void deal(int no, int add) {
  if (no != 1 && trans[no - 1] > trans[no]) ta.upd(no - 1, add);
  if (no != n && trans[no] > trans[no + 1]) ta.upd(no, add);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", orig + i);
    trans[orig[i]] = i;
  }
  for (int i = 1; i < n; i++)
    if (trans[i] > trans[i + 1]) ta.upd(i, 1);
  scanf("%d", &nq);
  for (int i = 0; i < nq; i++) {
    int cate, a, b;
    scanf("%d%d%d", &cate, &a, &b);
    if (cate == 1) {
      b--;
      int ans = ta.que(b);
      if (a > 1) ans -= ta.que(a - 1);
      printf("%d\n", ans + 1);
    } else {
      int x = orig[a], y = orig[b];
      swap(orig[a], orig[b]);
      if (x > y) swap(x, y);
      deal(x, -1);
      deal(y, -1);
      if (x == y - 1 && trans[x] > trans[y]) ta.upd(x, 1);
      swap(trans[x], trans[y]);
      deal(x, 1);
      deal(y, 1);
      if (x == y - 1 && trans[x] > trans[y]) ta.upd(x, -1);
    }
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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 double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[8][2] = {{1, 0}, {-1, 0}, {0, 1},  {0, -1},
                        {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a > b ? a : b;
  if (c != -1234567) ans = max1(ans, c);
  if (d != -1234567) ans = max1(ans, d);
  return ans;
}
template <class T>
inline T min1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a < b ? a : b;
  if (c != -1234567) ans = min(ans, c);
  if (d != -1234567) ans = min(ans, d);
  return ans;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
template <class T>
inline T lb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / abs1(a) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
inline int bitnum(unsigned long long nValue) {
  nValue = ((0xaaaaaaaaaaaaaaaaull & nValue) >> 1) +
           (0x5555555555555555ull & nValue);
  nValue = ((0xccccccccccccccccull & nValue) >> 2) +
           (0x3333333333333333ull & nValue);
  nValue = ((0xf0f0f0f0f0f0f0f0ull & nValue) >> 4) +
           (0x0f0f0f0f0f0f0f0full & nValue);
  nValue = ((0xff00ff00ff00ff00ull & nValue) >> 8) +
           (0x00ff00ff00ff00ffull & nValue);
  nValue = ((0xffff0000ffff0000ull & nValue) >> 16) +
           (0x0000ffff0000ffffull & nValue);
  nValue = ((0xffffffff00000000ull & nValue) >> 32) +
           (0x00000000ffffffffull & nValue);
  return nValue;
}
long long pow(long long n, long long m, long long mod = 0) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int maxn = 500100;
template <class t>
struct tree_array {
  t num[maxn], n, bigwei;
  tree_array(int nn = maxn) {
    n = nn;
    while (lb(nn) != nn) nn -= lb(nn);
    bigwei = nn;
  }
  void upd(int no, t add) {
    while (no <= n) {
      num[no] += add;
      no += lb(no);
    }
  }
  t que(int no) {
    t ans = 0;
    while (no) {
      ans += num[no];
      no -= lb(no);
    }
    return ans;
  }
  int getrank(t x) {
    int ans = 0, ranwei = bigwei;
    t rank = 0;
    while (ranwei) {
      if (rank + num[ranwei + ans] < x) {
        ans += ranwei;
        rank += num[ans];
      }
      ranwei >>= 1;
    }
    return ans + 1;
  }
};
tree_array<int> ta;
int n, orig[maxn];
int nq;
int trans[maxn];
inline void deal(int no, int add) {
  if (no != 1 && trans[no - 1] > trans[no]) ta.upd(no - 1, add);
  if (no != n && trans[no] > trans[no + 1]) ta.upd(no, add);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", orig + i);
    trans[orig[i]] = i;
  }
  for (int i = 1; i < n; i++)
    if (trans[i] > trans[i + 1]) ta.upd(i, 1);
  scanf("%d", &nq);
  for (int i = 0; i < nq; i++) {
    int cate, a, b;
    scanf("%d%d%d", &cate, &a, &b);
    if (cate == 1) {
      b--;
      int ans = ta.que(b);
      if (a > 1) ans -= ta.que(a - 1);
      printf("%d\n", ans + 1);
    } else {
      int x = orig[a], y = orig[b];
      swap(orig[a], orig[b]);
      if (x > y) swap(x, y);
      deal(x, -1);
      deal(y, -1);
      if (x == y - 1 && trans[x] > trans[y]) ta.upd(x, 1);
      swap(trans[x], trans[y]);
      deal(x, 1);
      deal(y, 1);
      if (x == y - 1 && trans[x] > trans[y]) ta.upd(x, -1);
    }
  }
  return 0;
}

```