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;
const int inf = 1.01e9;
const double eps = 1e-9;
int seed = 1;
int rnd() {
  int ret = seed;
  seed = (seed * (long long)7 + 13) % 1000000007;
  return ret;
}
set<pair<pair<int, int>, long long>> second;
int mpow(int a, int b, int mod) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * (long long)a) % mod;
    a = (a * (long long)a) % mod;
    b >>= 1;
  }
  return res;
}
long long a[(int)1e5 + 10];
int DEBUG = 1;
int main() {
  int n, m, vmax;
  scanf("%d%d%d%d", &n, &m, &seed, &vmax);
  for (int i = 0; (i) < (n); ++i) a[i] = rnd() % vmax + 1;
  for (int i = 0; (i) < (n); ++i) second.insert({{i, i + 1}, a[i]});
  for (int _ = 0; (_) < (m); ++_) {
    int op = rnd() % 4 + 1;
    int l = rnd() % n + 1;
    int r = rnd() % n + 1;
    if (l > r) swap(l, r);
    int x = -1, y = -1;
    if (op == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (op == 4) y = rnd() % vmax + 1;
    --l;
    if (op == 1) {
      auto it = second.upper_bound({{l, inf}, -1ll});
      it--;
      vector<pair<pair<int, int>, long long>> v;
      while (it != second.end() && it->first.first <= r) {
        v.push_back(*it);
        it++;
      }
      for (auto z : v) second.erase(z);
      for (auto z : v) {
        vector<int> pos = {z.first.first, z.first.second, l, r};
        sort((pos).begin(), pos.end());
        pos.resize(unique((pos).begin(), pos.end()) - pos.begin());
        for (int i = 0; (i) < (((int)(pos).size()) - 1); ++i) {
          int L = pos[i];
          int R = pos[i + 1];
          if (!(z.first.first <= L && R <= z.first.second)) continue;
          long long val = z.second;
          if (l <= L && R <= r) val += x;
          second.insert({{L, R}, val});
        }
      }
    }
    if (op == 2) {
      auto it = second.upper_bound({{l, inf}, -1ll});
      it--;
      vector<pair<pair<int, int>, long long>> v;
      while (it != second.end() && it->first.first <= r) {
        v.push_back(*it);
        it++;
      }
      for (auto z : v) second.erase(z);
      for (auto z : v) {
        vector<int> pos = {z.first.first, z.first.second, l, r};
        sort((pos).begin(), pos.end());
        pos.resize(unique((pos).begin(), pos.end()) - pos.begin());
        for (int i = 0; (i) < (((int)(pos).size()) - 1); ++i) {
          int L = pos[i];
          int R = pos[i + 1];
          if (!(z.first.first <= L && R <= z.first.second)) continue;
          long long val = z.second;
          if (l <= L && R <= r) {
            continue;
          }
          second.insert({{L, R}, val});
        }
      }
      second.insert({{l, r}, (long long)x});
    }
    if (op == 3) {
      auto it = second.upper_bound({{l, inf}, -1ll});
      it--;
      vector<pair<pair<int, int>, long long>> v;
      while (it != second.end() && it->first.first <= r) {
        v.push_back(*it);
        it++;
      }
      vector<pair<long long, int>> vv;
      for (auto z : v) {
        vector<int> pos = {z.first.first, z.first.second, l, r};
        sort((pos).begin(), pos.end());
        pos.resize(unique((pos).begin(), pos.end()) - pos.begin());
        for (int i = 0; (i) < (((int)(pos).size()) - 1); ++i) {
          int L = pos[i];
          int R = pos[i + 1];
          if (!(z.first.first <= L && R <= z.first.second)) continue;
          long long val = z.second;
          if (l <= L && R <= r) {
            vv.push_back({val, R - L});
          }
        }
      }
      sort((vv).begin(), vv.end());
      x--;
      int ok = 0;
      for (int i = 0; (i) < (((int)(vv).size())); ++i) {
        if (x < vv[i].second) {
          printf("%lld\n", vv[i].first);
          ok = 1;
          break;
        }
        x -= vv[i].second;
      }
      assert(ok);
    }
    if (op == 4) {
      auto it = second.upper_bound({{l, inf}, -1ll});
      it--;
      vector<pair<pair<int, int>, long long>> v;
      while (it != second.end() && it->first.first <= r) {
        v.push_back(*it);
        it++;
      }
      vector<pair<long long, int>> vv;
      for (auto z : v) {
        vector<int> pos = {z.first.first, z.first.second, l, r};
        sort((pos).begin(), pos.end());
        pos.resize(unique((pos).begin(), pos.end()) - pos.begin());
        for (int i = 0; (i) < (((int)(pos).size()) - 1); ++i) {
          int L = pos[i];
          int R = pos[i + 1];
          if (!(z.first.first <= L && R <= z.first.second)) continue;
          long long val = z.second;
          if (l <= L && R <= r) {
            vv.push_back({val, R - L});
          }
        }
      }
      int res = 0;
      for (int i = 0; (i) < (((int)(vv).size())); ++i) {
        res = (res + vv[i].second * (long long)mpow(vv[i].first % y, x, y)) % y;
      }
      printf("%d\n", res);
    }
  }
  return 0;
}