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 st {
  long long l, r, val;
  bool operator<(const st &o) const { return l < o.l; }
};
long long seed;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long binpow(long long x, long long y, long long m) {
  x %= m;
  long long ret = 1;
  while (y) {
    ret *= y & 1 ? x : 1;
    y /= 2, x *= x;
    ret %= m, x %= m;
  }
  return ret;
}
set<st> sh;
void dol(long long l) {
  for (auto &cv : sh) {
    if (cv.l <= l && cv.r >= l) {
      if (cv.l != l) {
        st nw;
        nw.val = cv.val;
        nw.l = l;
        nw.r = cv.r;
        st nw2 = cv;
        nw2.r = l - 1;
        sh.erase(cv);
        sh.insert(nw);
        sh.insert(nw2);
      }
      return;
    }
  }
}
void dor(long long r) {
  for (auto &cv : sh) {
    if (cv.l <= r && cv.r >= r) {
      if (cv.r != r) {
        st nw;
        nw.val = cv.val;
        nw.l = r + 1;
        nw.r = cv.r;
        st nw2 = cv;
        nw2.r = r;
        sh.erase(cv);
        sh.insert(nw);
        sh.insert(nw2);
      }
      return;
    }
  }
}
void prll() {
  cout << ("sh.size()") << " is " << (sh.size()) << endl;
  for (auto &cv : sh) cout << cv.l << ' ' << cv.r << ' ' << cv.val << endl;
}
void upd1(long long l, long long r, long long x) {
  dol(l), dor(r);
  vector<st> v;
  for (auto &cv : sh) {
    if (cv.l >= l && cv.r <= r) {
      v.push_back(cv);
    }
  }
  for (auto &cv : v) {
    sh.erase(cv);
    auto nw = cv;
    nw.val += x;
    sh.insert(nw);
  }
}
void upd2(long long l, long long r, long long x) {
  dol(l), dor(r);
  vector<st> v;
  for (auto &cv : sh) {
    if (cv.l >= l && cv.r <= r) {
      v.push_back(cv);
    }
  }
  for (auto &cv : v) sh.erase(cv);
  st nw;
  nw.l = l, nw.r = r, nw.val = x;
  sh.insert(nw);
}
long long query1(long long l, long long r, long long x) {
  dol(l), dor(r);
  vector<pair<long long, long long>> v;
  for (auto &cv : sh) {
    if (cv.l >= l && cv.r <= r) {
      v.push_back({cv.val, cv.r - cv.l + 1});
    }
  }
  sort(v.begin(), v.end());
  for (auto &cv : v) {
    if (cv.second >= x) return cv.first;
    x -= cv.second;
  }
}
long long query2(long long l, long long r, long long x, long long y) {
  dol(l), dor(r);
  long long ans = 0;
  for (auto &cv : sh) {
    if (cv.l >= l && cv.r <= r) {
      ans += (binpow(cv.val, x, y)) * (cv.r - cv.l + 1) % y;
      ans %= y;
    }
  }
  return ans;
}
void solve() {
  long long n, m, vmax;
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; i++) {
    long long x;
    x = rnd() % vmax + 1;
    st nw;
    nw.l = nw.r = i;
    nw.val = x;
    sh.insert(nw);
  }
  for (long long i = 0; i < m; i++) {
    long long t;
    t = rnd() % 4 + 1;
    long long l, r, x, y;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (t == 1) {
      x = (rnd() % vmax) + 1;
      upd1(l, r, x);
    } else if (t == 2) {
      x = (rnd() % vmax) + 1;
      upd2(l, r, x);
    } else if (t == 3) {
      x = (rnd() % (r - l + 1)) + 1;
      cout << query1(l, r, x) << '\n';
    } else {
      x = (rnd() % vmax) + 1;
      y = (rnd() % vmax) + 1;
      cout << query2(l, r, x, y) << '\n';
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}