Problem: — Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

<image>

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

  * 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
  * 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
  * 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
  * 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. <image>.

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:
    
    
      
    def rnd():  
      
        ret = seed  
        seed = (seed * 7 + 13) mod 1000000007  
        return ret  
      
    for i = 1 to n:  
      
        a[i] = (rnd() mod vmax) + 1  
      
    for i = 1 to m:  
      
        op = (rnd() mod 4) + 1  
        l = (rnd() mod n) + 1  
        r = (rnd() mod n) + 1  
      
        if (l > r):   
             swap(l, r)  
      
        if (op == 3):  
            x = (rnd() mod (r - l + 1)) + 1  
        else:  
            x = (rnd() mod vmax) + 1  
      
        if (op == 4):  
            y = (rnd() mod vmax) + 1  
      
    

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

Examples

Input

10 10 7 9


Output

2
1
0
3


Input

10 10 9 9


Output

1
1
3
3

Note

In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.

The operations are:

  * 2 6 7 9
  * 1 3 10 8
  * 4 4 6 2 4
  * 1 4 5 8
  * 2 1 7 1
  * 4 7 9 4 4
  * 1 2 7 9
  * 4 5 8 1 1
  * 2 5 7 5
  * 4 3 10 8 5

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.TreeMap;

/**
 * Created by mcl on 2017/12/2.
 */
public class Main {

  public static void main(String[] args) {
    new Main().cal();
  }

  static class Pair {
    long val;
    int cnt;

    Pair() {}
    Pair(long val, int cnt) {
      this.val = val;
      this.cnt = cnt;
    }
  }

  static class PairComparator implements Comparator<Pair> {
    @Override
    public int compare(Pair a, Pair b) {
      return a.val < b.val ? -1 : 1;
    }
  }

  int n;
  int m;
  int seed;
  int vmax;
  Pair[] b = null;
  Pair[] c = null;
  int bSize = 0;
  int cSize = 0;

  int rnd() {
    int ret = seed;
    seed = (int)((seed * 7L + 13) % 1000000007);
    return ret;
  }



  void cal() {
    Scanner in = new Scanner(System.in);
    n = in.nextInt();
    m = in.nextInt();
    seed = in.nextInt();
    vmax = in.nextInt();

    b = new Pair[n];
    c = new Pair[n];

    b[bSize ++] = new Pair(rnd() % vmax + 1, 1);
    for (int i = 2; i <= n; ++ i) {
      int val = rnd() % vmax + 1;
      if (val == b[bSize - 1].val) {
        b[bSize - 1].cnt += 1;
      }
      else {
        b[bSize ++] = new Pair(val, 1);
      }
    }

    for (int i = 0; i < m; ++ i) {
      int op = rnd() % 4 + 1;
      int left = rnd() % n + 1;
      int right = rnd() % n + 1;

      if (left > right) {
        int tmp = left;
        left = right;
        right = tmp;
      }

      int x = 0;
      int y = 0;

      if (op == 3) {
        x = rnd() % (right - left + 1) + 1;
      }
      else {
        x = rnd() % vmax + 1;
      }
      if (op == 4) {
        y = rnd() % vmax + 1;
      }

      if (op == 1 || op == 2) {
        operation1or2(left, right, x, op == 1);
      }
      else if (op == 3) {
        operation3(left, right, x);
      }
      else  {
        operation4(left, right, x, y);
      }
    }
  }

  void operation1or2(int left, int right, int x, boolean isOperation1) {
    int L = 0, R = 0;
    cSize = 0;
    for (int i = 0; i < bSize; ++ i) {
      L = R + 1;
      R = L + b[i].cnt - 1;
      add(b[i].val, intersect(L, left - 1, L, R));
      if (isOperation1) {
        add(b[i].val + x, intersect(left, right , L, R));
      }
      else {
        add(x, intersect(left, right , L, R));
      }

      add(b[i].val, intersect(right + 1, R, L, R));
    }
    for (int i = 0; i <  cSize; ++ i) {
      b[i] = c[i];
    }
    bSize = cSize;
  }
  void operation3(int left, int right, int kth) {
    int L = 0, R = 0;
    cSize = 0;
    for (int i = 0; i < bSize; ++ i) {
      L = R + 1;
      R = L + b[i].cnt - 1;
      int cnt = intersect(left, right , L, R);
      if (cnt > 0) {
        c[cSize ++] = new Pair(b[i].val, cnt);
      }
    }
    Arrays.sort(c, 0, cSize, new PairComparator());
    for (int i = 0; i < cSize; ++ i) {
      if (c[i].cnt >= kth) {
        System.out.println(c[i].val);
        return;
      }
      kth -= c[i].cnt;
    }

  }

  void operation4(int left, int right, int x, int y) {
    int L = 0, R = 0;

    long result = 0;

    for (int i = 0; i < bSize; ++ i) {
      L = R + 1;
      R = L + b[i].cnt - 1;
      int cnt = intersect(left, right , L, R);
      if (cnt > 0) {
        result += Pow(b[i].val, x, y) * cnt % y;
        result %= y;
      }
    }
    System.out.println(result);
  }

  int intersect(int left1, int right1, int left2, int right2) {
    int left = Math.max(left1, left2);
    int right = Math.min(right1, right2);
    return Math.max(0, right - left + 1);
  }

  void add(long val, int cnt) {
    if (cnt == 0) {
      return;
    }
    if (cSize > 0 && c[cSize - 1].val == val) {
      c[cSize - 1].cnt += cnt;
    }
    else {
      c[cSize ++] = new Pair(val, cnt);
    }
  }

  static long Pow(long a, int b, int mod) {
    long result = 1;
    a %= mod;
    while (b != 0) {
      if (b % 2 == 1) {
        result = result * a % mod;
      }
      a = a * a % mod;
      b >>= 1;
    }
    return result;
  }

}