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;
int add(int a, int b, int MOD) {
  if ((a += b) >= MOD) a -= MOD;
  return a;
}
int mul(int a, int b, int MOD) { return 1ll * a * b % MOD; }
int qpow(int a, int b, int MOD) {
  int r = 1;
  for (; b; b >>= 1) {
    if (b & 1) r = mul(r, a, MOD);
    a = mul(a, a, MOD);
  }
  return r;
}
const int N = 1e5 + 7;
int n, m, seed, vmax;
int a[N];
int rnd() {
  int ret = seed;
  seed = (7ll * seed + 13) % 1000000007;
  return ret;
}
vector<pair<long long, int> > v;
struct Nd {
  long long v, tval;
  char ttp;
  void chg(long long nv) {
    v = tval = nv;
    ttp = 'c';
  }
  void add(long long mv) {
    if (~v) v += mv;
    if (ttp != '-')
      tval += mv;
    else
      ttp = 'a', tval = mv;
  }
  void upd(const Nd &ls, const Nd &rs) {
    ttp = '-';
    v = ls.v == rs.v ? ls.v : -1;
  }
};
struct Tree {
  static const int N = ::N << 2;
  Nd nd[N];
  void upd(int o) { nd[o].upd(nd[((o) << 1)], nd[((o) << 1 | 1)]); }
  void dwn(int o) {
    if (nd[o].ttp == '-') return;
    if (nd[o].ttp == 'a') {
      nd[((o) << 1)].add(nd[o].tval);
      nd[((o) << 1 | 1)].add(nd[o].tval);
    } else {
      nd[((o) << 1)].chg(nd[o].tval);
      nd[((o) << 1 | 1)].chg(nd[o].tval);
    }
    nd[o].ttp = '-';
  }
  void bud(int o, int l, int r) {
    if (l == r) {
      nd[o].v = a[l];
      nd[o].ttp = '-';
      return;
    }
    int mid = l + r >> 1;
    bud(((o) << 1), l, mid);
    bud(((o) << 1 | 1), mid + 1, r);
    upd(o);
  }
  void add(int o, int l, int r, int ql, int qr, long long qv) {
    if (ql <= l && qr >= r) {
      nd[o].add(qv);
      return;
    }
    dwn(o);
    int mid = l + r >> 1;
    if (ql <= mid) add(((o) << 1), l, mid, ql, qr, qv);
    if (qr > mid) add(((o) << 1 | 1), mid + 1, r, ql, qr, qv);
    upd(o);
  }
  void chg(int o, int l, int r, int ql, int qr, long long qv) {
    if (ql <= l && qr >= r) {
      nd[o].chg(qv);
      return;
    }
    dwn(o);
    int mid = l + r >> 1;
    if (ql <= mid) chg(((o) << 1), l, mid, ql, qr, qv);
    if (qr > mid) chg(((o) << 1 | 1), mid + 1, r, ql, qr, qv);
    upd(o);
  }
  void qry(int o, int l, int r, int ql, int qr) {
    if (ql <= l && qr >= r && ~nd[o].v) {
      v.push_back({nd[o].v, r - l + 1});
      return;
    }
    dwn(o);
    int mid = l + r >> 1;
    if (ql <= mid) qry(((o) << 1), l, mid, ql, qr);
    if (qr > mid) qry(((o) << 1 | 1), mid + 1, r, ql, qr);
    upd(o);
  }
} tree;
void solve() {
  for (int i = (1); i < (n + 1); ++i) a[i] = rnd() % vmax + 1;
  tree.bud(1, 1, n);
  for (int i = (1); i < (m + 1); ++i) {
    int op = (rnd() % 4) + 1;
    int l = (rnd() % n) + 1;
    int r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    int x = op == 3 ? (rnd() % (r - l + 1)) + 1 : (rnd() % vmax) + 1;
    int y;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1) {
      tree.add(1, 1, n, l, r, x);
    } else if (op == 2) {
      tree.chg(1, 1, n, l, r, x);
    } else if (op == 3) {
      v.clear();
      tree.qry(1, 1, n, l, r);
      sort((v).begin(), (v).end());
      for (int i = (0); i < (((int)(v).size())); ++i) {
        if (x <= v[i].second) {
          printf("%lld\n", v[i].first);
          break;
        } else
          x -= v[i].second;
      }
    } else {
      v.clear();
      tree.qry(1, 1, n, l, r);
      int ans = 0;
      for (int l = 0, r; l < ((int)(v).size()); l = r) {
        int cnt = 0;
        r = l;
        while (r < ((int)(v).size()) && v[r].first == v[l].first)
          cnt += v[r++].second;
        ans = add(ans, mul(qpow(v[l].first % y, x, y), cnt, y), y);
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  while (~scanf("%d%d%d%d", &n, &m, &seed, &vmax)) solve();
  return 0;
}