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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
const long long gmod = 1000000007;
map<int, pair<long long, long long> > intervals;
long long seed;
long long n, m;
long long vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % gmod;
  return ret;
}
void setInt(int beg, int end, long long val) {
  map<int, pair<long long, long long> >::iterator it =
      intervals.upper_bound(beg);
  it--;
  if (it->first < beg) {
    if (it->second.first > end) {
      intervals[end + 1] = it->second;
    }
    it->second.first = beg - 1;
    it++;
  }
  while (it != intervals.end()) {
    if (it->first > end) break;
    if (it->second.first <= end) {
      map<int, pair<long long, long long> >::iterator it2 = it;
      it++;
      intervals.erase(it2);
      continue;
    }
    intervals[end + 1] = it->second;
    map<int, pair<long long, long long> >::iterator it2 = it;
    it++;
    intervals.erase(it2);
  }
  intervals[beg] = pair<long long, long long>(end, val);
}
void addVal(int beg, int end, long long val) {
  map<int, pair<long long, long long> >::iterator it =
      intervals.upper_bound(beg);
  it--;
  if (it->first < beg) {
    intervals[beg] = it->second;
    it->second.first = beg - 1;
    it++;
  }
  while (it != intervals.end()) {
    if (it->first > end) break;
    if (it->second.first <= end) {
      it->second.second += val;
      it++;
      continue;
    }
    intervals[end + 1] = it->second;
    it->second.first = end;
    it->second.second += val;
    it++;
  }
}
pair<long long, long long> have[110000];
int haveCount;
long long smallest(int beg, int end, int x) {
  haveCount = 0;
  map<int, pair<long long, long long> >::iterator it =
      intervals.upper_bound(beg);
  it--;
  while (it != intervals.end() && it->first <= end) {
    int curBeg = max(beg, it->first);
    int curEnd = min((long long)end, it->second.first);
    int len = curEnd - curBeg + 1;
    if (len <= 0) {
      cout << "NOO" << endl;
      exit(1);
    }
    have[haveCount++] = pair<long long, long long>(it->second.second, len);
    it++;
  }
  sort(have, have + haveCount);
  for (int i = 0; i < (haveCount); i++) {
    x -= have[i].second;
    if (x <= 0) return have[i].first;
  }
  exit(1);
}
long long power(long long a, long long b, long long mod) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long powerSum(int beg, int end, long long x, long long y) {
  long long res = 0;
  map<int, pair<long long, long long> >::iterator it =
      intervals.upper_bound(beg);
  it--;
  while (it != intervals.end() && it->first <= end) {
    int curBeg = max(beg, it->first);
    int curEnd = min((long long)end, it->second.first);
    int len = curEnd - curBeg + 1;
    res = (res + power(it->second.second, x, y) * len) % y;
    it++;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= (n); i++) {
    intervals[i] = pair<long long, long long>(i, rnd() % vmax + 1);
  }
  for (int i = 0; i < (m); i++) {
    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;
    if (op == 1) {
      addVal(l, r, x);
    } else if (op == 2) {
      setInt(l, r, x);
    } else if (op == 3) {
      long long ans = smallest(l, r, x);
      cout << ans << endl;
    } else if (op == 4) {
      long long ans = powerSum(l, r, x, y);
      cout << ans << endl;
    } else
      int asf = 15;
  }
  return 0;
}