RT，线段树维护区间和，平方和，支持区间加和区间对固定数取模（这两个操作其实是固定在一起的，相当于 $a_i \to (a_i + k)\%p$），应该可以做到 $O(nlog_n)$ 吧？

但是蒟蒻的算法严重超时了，非常不解。

#pragma GCC optimized(2)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
inline void read(int &x)
{
	x = 0;
	int f = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9')
	{
		f |= ch == '-';
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9')
	{
		x = x * 10 + ch - 48;
		ch = getchar();
	}
	x = f ? -x : x;
	return;
}
#define N 500005
int n, q, p, num[N];
struct node
{
	int l, r, val, maxx, tag;
	ll sum;
};
class segment_tree
{
	public:
	node tree[N << 2];
	inline void pushup(int root)
	{
		tree[root].val = tree[root << 1].val + tree[root << 1 | 1].val;
		tree[root].sum = tree[root << 1].sum + tree[root << 1 | 1].sum;
		tree[root].maxx = max(tree[root << 1].maxx, tree[root << 1 | 1].maxx);
		return;
	}
	void build(int root, int l, int r)
	{
		tree[root].l = l;
		tree[root].r = r;
		if(l == r)
		{
			tree[root].val = tree[root].maxx = num[l];
			tree[root].sum = 1ll * num[l] * num[l];
			return;
		}
		int mid = (l + r) >> 1;
		build(root << 1, l, mid);
		build(root << 1 | 1, mid + 1, r);
		pushup(root);
		return;
	}
	void addtag(int root, int v)
	{
		tree[root].maxx += v;
		tree[root].sum += 1ll * (tree[root].r - tree[root].l + 1) * v * v + 1ll * 2 * tree[root].val * v;
		tree[root].val += (tree[root].r - tree[root].l + 1) * v;
		tree[root].tag += v;
		return;
	}
	void pushdown(int root)
	{
		if(tree[root].tag)
		{
			addtag(root << 1, tree[root].tag);
			addtag(root << 1 | 1, tree[root].tag);
			tree[root].tag = 0;
		}
		return;
	}
	void updateadd(int root, int L, int R, int v)
	{
		if(L <= tree[root].l && tree[root].r <= R)
		{
			addtag(root, v);
			return;
		}
		pushdown(root);
		int mid = (tree[root].l + tree[root].r) >> 1;
		if(L <= mid)
		{
			updateadd(root << 1, L, R, v);
		}
		if(mid < R)
		{
			updateadd(root << 1 | 1, L, R, v);
		}
		pushup(root);
		return;
	}
	void updatemod(int root, int L, int R)
	{
		if(tree[root].l == tree[root].r)
		{
			tree[root].val %= p;
			tree[root].sum = 1ll * tree[root].val * tree[root].val;
			tree[root].maxx = tree[root].val;
			return;
		}
		pushdown(root);
		if(tree[root].maxx < p)
		{
			return;
		}
		int mid = (tree[root].l + tree[root].r) >> 1;
		if(L <= mid && tree[root << 1].maxx >= p)
		{
			updatemod(root << 1, L, R);
		}
		if(mid < R && tree[root << 1 | 1].maxx >= p)
		{
			updatemod(root << 1 | 1, L, R);
		}
		pushup(root);
		return;
	}
	pair<int, ll> query(int root, int L, int R)
	{
		if(L <= tree[root].l && tree[root].r <= R)
		{
			return make_pair(tree[root].val, tree[root].sum);
		}
		pushdown(root);
		int mid = (tree[root].l + tree[root].r) >> 1;
		pair<int, ll> tmp, ret = make_pair(0, 0ll);
		if(L <= mid)
		{
			tmp = query(root << 1, L, R);
			ret.first += tmp.first, ret.second += tmp.second;
		}
		if(mid < R)
		{
			tmp = query(root << 1 | 1, L, R);
			ret.first += tmp.first, ret.second += tmp.second;
		}
		return ret;
	}
};
segment_tree T;
signed main()
{
//	freopen("pockets.in", "r", stdin);
//	freopen("pockets.out", "w", stdout);
	srand(time(0));
	std::mt19937 rng(rand());
    std::uniform_int_distribution<int> rd(1, 1e9);
	read(n), read(q), read(p);
	for(int i = 1; i <= n; i++)
	{
		read(num[i]);
	}
	T.build(1, 1, n);
	int opt, x, y, v, len;
	while(q--)
	{
		read(opt);
		switch(opt)
		{
			case 1:
			{
				read(x), read(y), read(v);
				T.updateadd(1, x, y, v);
				T.updatemod(1, x, y);
				break;
			}
			case 2:
			{
				read(x), read(y), read(v);
				len = rd(rng) % v;
				if((T.query(1, x, x + len) == T.query(1, y, y + len)) && (T.query(1, x + len, x + v - 1) == T.query(1, y + len, y + v - 1)))
				{
					printf("ye5\n");
				}
				else
				{
					printf("n0\n");
				}
				break;
			}
			default:
			{
				return -1;
			}
		}
	}
	return 0;
}