#include<iostream>
#include<cstdio>
#include<algorithm>
#define rep(i, a, b) for(int i = (a); i <= (int)(b); i++)
#define per(i, a, b) for(int i = (a); i >= (int)(b); i--)
using namespace std;
typedef long long ll;
const int N = 200010;
int m, k;
struct node{    // m 位 k 进制的数
    int a[30];
    node() {rep(i, 0, m - 1) a[i] = 0;}
    node(int x) {rep(i, 0, m - 1) a[i] = x % k, x /= k;}
    int& operator [] (const int &x) {return a[x];}
    int operator [] (const int &x) const {return a[x];}
    int val(){
        int ret = 0;
        per(i, m - 1, 0) ret = ret * k + a[i];
        return ret;
    }
};
node operator + (const node &x, const node &y){
    node ret;
    rep(i, 0, m - 1) ret[i] = (x[i] + y[i]) % k;
    return ret;
}
node operator - (const node &x, const node &y){
    node ret;
    rep(i, 0, m - 1) ret[i] = (x[i] - y[i] + k) % k;
    return ret;
}
node operator * (const node &x, const int &y){
    node ret = x;
    rep(i, 0, m - 1) ret[i] = (ll)ret[i] * y % k;
    return ret;
}

void exgcd(ll a, ll b, ll &x, ll &y){
    if(!b) {x = 1, y = 0; return;}
    exgcd(b, a % b, y, x);
    y -= a / b * x;
}

ll inv(ll a) {ll x = 0, y = 0; exgcd(a, k, x, y); return (x % k + k) % k;}

node b[30];
void insert(node x){
    per(i, m - 1, 0) if(x[i]){
        if(!b[i][i]) b[i] = x * inv(x[i]), x = x * (k / __gcd(x[i], k));
        else while(x[i]){
            b[i] = b[i] - x * (b[i][i] / x[i]);
            swap(x, b[i]);
        }
    }
}

node query(node x){
    per(i, m - 1, 0) if(b[i][i])
        x = x - b[i] * (x[i] / b[i][i]);
    return x;
}

int n = 1;
node dis[N];

int q;
int main(){
    scanf("%d%d%d", &q, &k, &m);
    while(q--){
        int op, x, y, z; scanf("%d", &op);
        if(op == 1){
            scanf("%d%d", &x, &y);
            dis[++n] = dis[x] + y;
            insert((node)y * 2);
        }
        else if(op == 2){
            scanf("%d%d%d", &x, &y, &z);
            insert((node)z * 2);
            insert(dis[x] + dis[y] + z);
        }
        else{
            scanf("%d%d", &x, &y);
            int ans = query(dis[x] + dis[y]).val();
            printf("%d\n", ans);
        }
    }
    return 0;
}