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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Data {
  long long st, dr, val;
  bool operator<(const Data& other) const { return st < other.st; }
} v[100005];
set<Data> s;
struct Sort {
  long long nr, val;
  bool operator<(const Sort& other) const { return val < other.val; }
} v1[100005];
long long seed;
long long rnd() {
  long long ret = seed;
  seed = (1LL * seed * 7 + 13) % 1000000007;
  return ret;
}
long long myPow(long long a, long long b, long long mod) {
  long long ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = 1LL * ans * a % mod;
    a = 1LL * a * a % mod;
  }
  return ans;
}
int main() {
  long long n, m, vmax;
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; ++i) {
    long long x = (rnd() % vmax) + 1;
    s.insert({i, i, x});
  }
  for (long long j = 1; j <= m; ++j) {
    long long op, l, r, x, y;
    op = rnd() % 4 + 1;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) {
      y = rnd() % vmax + 1;
    }
    long long a = l, b = r;
    set<Data>::iterator it = s.lower_bound({a, 0, 0});
    long long p = 0;
    while (it != s.end() && it->dr <= b) {
      v[++p] = {it->st, it->dr, it->val};
      s.erase(it);
      it = s.lower_bound({a, 0, 0});
    }
    it = s.lower_bound({a, 0, 0});
    if (it != s.end() && it->st <= b) {
      v[++p] = {it->st, b, it->val};
      long long dr1 = it->dr;
      long long val1 = it->val;
      s.erase(it);
      s.insert({b + 1, dr1, val1});
    }
    it = s.lower_bound({a, 0, 0});
    if (it != s.begin()) {
      it--;
      if (it->dr >= a) {
        v[++p] = {a, min(it->dr, b), it->val};
        long long st1 = it->st;
        long long dr1 = it->dr;
        long long val1 = it->val;
        s.erase(it);
        s.insert({st1, a - 1, val1});
        if (b < dr1) s.insert({b + 1, dr1, val1});
      }
    }
    if (op == 2) {
      s.insert({l, r, x});
    } else if (op == 1) {
      for (long long i = 1; i <= p; ++i)
        s.insert({v[i].st, v[i].dr, v[i].val + x});
    } else if (op == 3) {
      for (long long i = 1; i <= p; ++i) {
        v1[i] = {v[i].dr - v[i].st + 1, v[i].val};
        s.insert(v[i]);
      }
      sort(v1 + 1, v1 + p + 1);
      long long nr = 0, ind = 1, val;
      while (ind <= p && nr < x) {
        val = v1[ind].val;
        nr += v1[ind].nr;
        ind++;
      }
      printf("%I64d\n", val);
    } else {
      long long ans = 0;
      for (long long i = 1; i <= p; ++i) {
        ans = (1LL * ans +
               (v[i].dr - v[i].st + 1) * myPow(v[i].val % y, x, y) % y) %
              y;
        s.insert(v[i]);
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}