Codecraft-18 and Codeforces Round #458 D. Bash and a Tough Math Puzzle(线段树)

D. Bash and a Tough Math Puzzle

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bash likes playing with arrays. He has an array a1, a2, ... an of n integers. He likes to guess the greatest common divisor (gcd) of different segments of the array. Of course, sometimes the guess is not correct. However, Bash will be satisfied if his guess is almost correct.

Suppose he guesses that the gcd of the elements in the range [l, r] of a is x. He considers the guess to be almost correct if he can change at most one element in the segment such that the gcd of the segment is x after making the change. Note that when he guesses, he doesn't actually change the array — he just wonders if the gcd of the segment can be made x. Apart from this, he also sometimes makes changes to the array itself.

Since he can't figure it out himself, Bash wants you to tell him which of his guesses are almost correct. Formally, you have to process qqueries of one of the following forms:

• 1 l r x — Bash guesses that the gcd of the range [l, r] is x. Report if this guess is almost correct.
• 2 i y — Bash sets ai to y.

Note: The array is 1-indexed.

Input

The first line contains an integer n (1 ≤ n ≤ 5·105)  — the size of the array.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109)  — the elements of the array.

The third line contains an integer q (1 ≤ q ≤ 4·105)  — the number of queries.

The next q lines describe the queries and may have one of the following forms:

• 1 l r x (1 ≤ l ≤ r ≤ n, 1 ≤ x ≤ 109).
• 2 i y (1 ≤ i ≤ n, 1 ≤ y ≤ 109).

Guaranteed, that there is at least one query of first type.

Output

For each query of first type, output "YES" (without quotes) if Bash's guess is almost correct and "NO" (without quotes) otherwise.

Examples

input

3
2 6 3
4
1 1 2 2
1 1 3 3
2 1 9
1 1 3 2

output

YES
YES
NO

input

5
1 2 3 4 5
6
1 1 4 2
2 3 6
1 1 4 2
1 1 5 2
2 5 10
1 1 5 2

output

NO
YES
NO
YES

Note

In the first sample, the array initially is {2, 6, 3}.

For query 1, the first two numbers already have their gcd as 2.

For query 2, we can achieve a gcd of 3 by changing the first element of the array to 3. Note that the changes made during queries of type 1are temporary and do not get reflected in the array.

After query 3, the array is now {9, 6, 3}.

For query 4, no matter which element you change, you cannot get the gcd of the range to be 2.

#include 
using namespace std;
int a[500005];
struct tree{
	int l, r, v;
}c[500005 * 3];
vectorg;
int gcd(int a, int b){
	return b == 0 ? a : gcd(b, a % b);
}
void build(int o, int l, int r){
	c[o].l = l; c[o].r = r;
	if(l == r){
		c[o].v = a[l]; return;
	}
	int mid = l + r >> 1;
	build(o << 1, l, mid);
	build(o << 1 | 1, mid + 1, r);
	c[o].v = gcd(c[o << 1].v, c[o << 1 | 1].v);
}
void update(int o, int pos, int v){
	if(c[o].l == c[o].r){
		c[o].v = v; return;
	}
	int mid = c[o].l + c[o].r >> 1;
	if(pos <= mid) update(o << 1, pos, v);
	else update(o << 1 | 1, pos, v);
	c[o].v = gcd(c[o << 1].v, c[o << 1 | 1].v);
}
void query(int o, int L, int R){
	if(c[o].l >= L && c[o].r <= R){
		g.push_back(o); return;
	}
	int mid = c[o].l + c[o].r >> 1;
	if(mid >= L) query(o << 1, L, R);
	if(mid < R) query(o << 1 | 1, L, R);
}
bool check(int o, int v){
	if(c[o].l == c[o].r){
		return 1;
	}
	int mid = c[o].l + c[o].r >> 1;
	if(c[o << 1].v % v != 0 && c[o << 1 | 1].v % v != 0){
		return false;
	}
	if(c[o << 1].l == c[o << 1].r){
		if(c[o << 1].v % v == 0 || c[o << 1 | 1].v % v == 0){
			return true;
		}
		else{
			return false;
		}
	}
	if(c[o << 1].v % v != 0){
		return check(o << 1, v);
	}
	else{
		return check(o << 1 | 1, v);
	}
}
int main(){
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%d", &a[i]);
	}
	build(1, 1, n);
	int q, op, x, y, v, l, r, mid;;
	scanf("%d", &q);
	while(q--){
		scanf("%d", &op);
		if(op == 2){
			scanf("%d %d", &x, &y);
			update(1, x, y);
		}
		if(op == 1){
			scanf("%d %d %d", &x, &y, &v);
			g.clear();
			query(1, x, y);
			int cnt = 0, pos;
			for(int i = 0; i < g.size(); ++i){
				if(c[g[i]].v % v != 0){
					cnt++;
					pos = g[i];
				}
			}
			if(cnt == 0) printf("YES\n");
			if(cnt > 1) printf("NO\n");
			if(cnt == 1){
				if(check(pos, v)) printf("YES\n");
				else printf("NO\n");
			}
		}
	}
}

/*
题意：
5e5个数，4e5次操作，每次要么询问一个区间[l,r]内所有数的GCD是否为v(可以改变区间内的一个数，不影响后面操作),
要么修改一个数。

思路：
线段树维护区间gcd，然后分类讨论一下。对于允许一次的修改操作，如果需要修改，我们一定把这个数改为v，对吧。
那么只要看分治整个区间，如果左右子区间gcd都不能整出v，那么肯定NO，否则就递归下去。
*/