You are given two permutations a and b, both consisting of n elements. Permutation of n elements is such a integer sequence that each value from 1 to n appears exactly once in it.

You are asked to perform two types of queries with them:

  * 1~l_a~r_a~l_b~r_b — calculate the number of values which appear in both segment [l_a; r_a] of positions in permutation a and segment [l_b; r_b] of positions in permutation b; 
  * 2~x~y — swap values on positions x and y in permutation b. 



Print the answer for each query of the first type.

It is guaranteed that there will be at least one query of the first type in the input.

Input

The first line contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5) — the number of elements in both permutations and the number of queries.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — permutation a. It is guaranteed that each value from 1 to n appears in a exactly once.

The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n) — permutation b. It is guaranteed that each value from 1 to n appears in b exactly once.

Each of the next m lines contains the description of a certain query. These are either:

  * 1~l_a~r_a~l_b~r_b (1 ≤ l_a ≤ r_a ≤ n, 1 ≤ l_b ≤ r_b ≤ n); 
  * 2~x~y (1 ≤ x, y ≤ n, x ≠ y). 

Output

Print the answers for the queries of the first type, each answer in the new line — the number of values which appear in both segment [l_a; r_a] of positions in permutation a and segment [l_b; r_b] of positions in permutation b.

Example

Input


6 7
5 1 4 2 3 6
2 5 3 1 4 6
1 1 2 4 5
2 2 4
1 1 2 4 5
1 2 3 3 5
1 1 6 1 2
2 4 1
1 4 4 1 3


Output


1
1
1
2
0

Note

Consider the first query of the first example. Values on positions [1; 2] of a are [5, 1] and values on positions [4; 5] of b are [1, 4]. Only value 1 appears in both segments.

After the first swap (the second query) permutation b becomes [2, 1, 3, 5, 4, 6].

After the second swap (the sixth query) permutation b becomes [5, 1, 3, 2, 4, 6].
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse4.2,sse4.1,ssse3,sse3,sse2,sse")
#pragma GCC target("avx")
#pragma GCC target("avx2")
using namespace std;
const int maxn = (int)2e5;
alignas(32) int a[maxn];
alignas(32) int b[maxn];
alignas(32) int posA[maxn];
alignas(32) int posB[maxn];
alignas(32) int special[maxn];
alignas(32) int result[128];
bool check(int lb, int rb, int i) {
  return lb <= special[i] && special[i] <= rb;
}
int getCount(int la, int ra, int lb, int rb) {
  int count = 0;
  if (la / 128 == ra / 128) {
    for (int i = la; i <= ra; i++) count += check(lb, rb, i);
    return count;
  } else {
    while (la % 128 != 0) {
      count += check(lb, rb, la);
      la++;
    }
    while (ra % 128 != 128 - 1) {
      count += check(lb, rb, ra);
      ra--;
    }
    ra++;
    for (int i = 0; i < 128; i++) result[i] = 0;
    for (int i = la; i + 128 <= ra; i += 128) {
      for (int j = 0; j < 128; j++) {
        result[j] += check(lb, rb, i + j);
      }
    }
    for (int i = 0; i < 128; i++) count += result[i];
  }
  return count;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i], a[i]--;
  for (int i = 0; i < n; i++) cin >> b[i], b[i]--;
  for (int i = 0; i < n; i++) posA[a[i]] = i;
  for (int i = 0; i < n; i++) posB[b[i]] = i;
  for (int i = 0; i < n; i++) special[i] = posB[a[i]];
  for (int i = 0; i < m; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int la, ra, lb, rb;
      cin >> la >> ra >> lb >> rb;
      la--, ra--, lb--, rb--;
      int count = getCount(la, ra, lb, rb);
      cout << count << '\n';
    } else {
      int x, y;
      cin >> x >> y;
      x--, y--;
      swap(b[x], b[y]);
      posB[b[x]] = x;
      posB[b[y]] = y;
      special[posA[b[x]]] = x;
      special[posA[b[y]]] = y;
    }
  }
}
