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 long long maxn = 100000 + 5;
long long a[maxn];
long long n, m;
long long seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long mn[maxn << 2], mx[maxn << 2], lazy[maxn << 2];
void pushup(long long rt) {
  mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
  mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
}
void pushdown(long long rt) {
  if (lazy[rt]) {
    mn[rt << 1] = mx[rt << 1] = lazy[rt << 1] = lazy[rt];
    mn[rt << 1 | 1] = mx[rt << 1 | 1] = lazy[rt << 1 | 1] = lazy[rt];
    lazy[rt] = 0;
  }
}
void build(long long l, long long r, long long rt) {
  if (l == r) {
    mn[rt] = mx[rt] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushup(rt);
}
void op1(long long L, long long R, long long l, long long r, long long rt,
         long long x) {
  if (L <= l && r <= R && mx[rt] == mn[rt]) {
    mx[rt] = mn[rt] = lazy[rt] = x + mn[rt];
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(rt);
  if (L <= mid) {
    op1(L, R, l, mid, rt << 1, x);
  }
  if (R > mid) {
    op1(L, R, mid + 1, r, rt << 1 | 1, x);
  }
  pushup(rt);
}
void op2(long long L, long long R, long long l, long long r, long long rt,
         long long x) {
  if (L <= l && r <= R) {
    mx[rt] = mn[rt] = lazy[rt] = x;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(rt);
  if (L <= mid) {
    op2(L, R, l, mid, rt << 1, x);
  }
  if (R > mid) {
    op2(L, R, mid + 1, r, rt << 1 | 1, x);
  }
  pushup(rt);
}
vector<pair<long long, long long> > v;
void get(long long L, long long R, long long l, long long r, long long rt) {
  if (L <= l && R >= r && mn[rt] == mx[rt]) {
    v.push_back(make_pair(mn[rt], r - l + 1));
    return;
  }
  pushdown(rt);
  long long mid = (l + r) >> 1;
  if (L <= mid) {
    get(L, R, l, mid, rt << 1);
  }
  if (R > mid) {
    get(L, R, mid + 1, r, rt << 1 | 1);
  }
}
long long op3(long long l, long long r, long long x) {
  v.clear();
  get(l, r, 1, n, 1);
  sort(v.begin(), v.end());
  for (size_t i = 0; i < v.size(); i++) {
    if (x <= v[i].second) {
      return v[i].first;
    }
    x -= v[i].second;
  }
}
long long pow_mod(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
long long op4(long long l, long long r, long long x, long long y) {
  v.clear();
  get(l, r, 1, n, 1);
  sort(v.begin(), v.end());
  long long preFirst = 0, preSecond = 0, res = 0;
  for (size_t i = 0; i < v.size(); i++) {
    if (v[i].first != preFirst) {
      res = (res + preSecond * pow_mod(preFirst % y, x, y) % y) % y;
      preFirst = v[i].first;
      preSecond = 0;
    }
    preSecond += v[i].second;
  }
  return (res + preSecond * pow_mod(preFirst % y, x, y) % y) % y;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
  }
  build(1, n, 1);
  for (long long i = 1; i <= m; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    if (l > r) {
      swap(l, r);
    }
    long long x, y;
    if (op == 3) {
      x = (rnd() % (r - l + 1)) + 1;
    } else {
      x = (rnd() % vmax) + 1;
    }
    if (op == 4) {
      y = (rnd() % vmax) + 1;
    }
    if (op == 1) {
      op1(l, r, 1, n, 1, x);
    } else if (op == 2) {
      op2(l, r, 1, n, 1, x);
    } else if (op == 3) {
      cout << op3(l, r, x) << endl;
    } else {
      cout << op4(l, r, x, y) << endl;
    }
  }
  return 0;
}