#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll powmod(ll a, ll e) {
  ll sum = 1;
  ll cur = a;
  while (e > 0) {
    if (e % 2) {
      sum = sum * cur % mod;
    }
    cur = cur * cur % mod;
    e /= 2;
  }
  return sum;
}

/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid M. Note that constructing this tree requires the identity
 * element of M and the operation of M.
 * Header requirement: vector, algorithm
 * Verified by AtCoder ABC017-D (http://abc017.contest.atcoder.jp/submissions/660402)
 */
template<class I, class BiOp = I (*) (I, I)>
class SegTree {
  int n;
  std::vector<I> dat;
  BiOp op;
  I e;
public:
  SegTree(int n_, BiOp op, I e) : op(op), e(e) {
    n = 1;
    while (n < n_) { n *= 2; } // n is a power of 2
    dat.resize(2 * n);
    for (int i = 0; i < 2 * n - 1; i++) {
      dat[i] = e;
    }
  }
  /* ary[k] <- v */
  void update(int k, I v) {
    k += n - 1;
    dat[k] = v;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = op(dat[2 * k + 1], dat[2 * k + 2]);
    }
  }
  void update_array(int k, int len, const I *vals) {
    for (int i = 0; i < len; ++i) {
      update(k + i, vals[i]);
    }
  }
  /*
    Updates all elements. O(n)
   */
  void update_all(const I *vals, int len) {
    for (int k = 0; k < std::min(n, len); ++k) {
      dat[k + n - 1] = vals[k];
    }
    for (int k = std::min(n, len); k < n; ++k) {
      dat[k + n - 1] = e;
    }
    for (int b = n / 2; b >= 1; b /= 2) {
      for (int k = 0; k < b; ++k) {
	dat[k + b - 1] = op(dat[k * 2 + b * 2 - 1], dat[k * 2 + b * 2]);
      }
    }
  }
  /* l,r are for simplicity */
  I querySub(int a, int b, int k, int l, int r) const {
    // [a,b) and  [l,r) intersects?
    if (r <= a || b <= l) return e;
    if (a <= l && r <= b) return dat[k];
    I vl = querySub(a, b, 2 * k + 1, l, (l + r) / 2);
    I vr = querySub(a, b, 2 * k + 2, (l + r) / 2, r);
    return op(vl, vr);
  }
  /* [a, b] (note: inclusive) */
  I query(int a, int b) const {
    return querySub(a, b + 1, 0, 0, n);
  }
};

struct mp {
    ll operator()(ll a, ll b) const {
        return (a + b) % mod;
    }
};

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n, q;
    ll a, b;
    cin >>n >> a >> b >> q;
    SegTree<ll, mp> st(n, mp(), 0);
    vector<ll> ary(n);
    vector<ll> pw(n);
    ll ba = b * powmod(a , mod - 2) % mod;
    ba = (mod - ba) % mod;
    pw[0] = 1;
    for (int i = 1; i < n; ++i) {
        pw[i] = pw[i - 1] * ba % mod;
    }
    for (int i = 0; i < n; ++i) {
        cin >> ary[i];
        st.update(i, ary[i] * pw[i] % mod);
    }
    while (q--) {
        int ty, l, r;
        cin >> ty >> l >> r;
        if (ty == 1) {
            st.update(l, r * pw[l] % mod);
            ary[l] = r;
        } else {
            if (ba == 0) {
                ll t = ary[l];
                cout << (t == 0 ? "Yes" : "No") << endl;
            } else {
                cout << (st.query(l ,r) == 0 ? "Yes" : "No") << endl;
            }
        }
    }
    return 0;
}