codeforces 551 E. GukiZ and GukiZiana （分块）

E. GukiZ and GukiZiana

time limit per test

10 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:

1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
2. Second type has form 2 y and asks you to find value of GukiZiana(a, y).

For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

Examples

input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3

output

2

input

2 3
1 2
1 2 2 1
2 3
2 4

output

0
-1

题解：分块

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define N 500003
#define LL long long
using namespace std;
int belong[N],l[N],r[N];
LL val[N],delta[N];
struct point
{
	LL x; int pos;
};
struct data{
	point a[720];
}c[720];
int n,m;
int cmp(point a,point b)
{
	return a.x<b.x||a.x==b.x&&a.pos<b.pos;
}
void build(int x)
{
	int cnt=0;
	for (int i=l[x];i<=r[x];i++) c[x].a[++cnt].x=val[i],c[x].a[cnt].pos=i;
	sort(c[x].a+1,c[x].a+cnt+1,cmp);
}
void calc(int &x,int &y,int k,LL v)
{
	LL vl=v-delta[k];
	if (vl<=0) return;
	int ls=1; int rs=r[k]-l[k]+1; int ans=n+1;
	while (ls<=rs) {
		int mid=(ls+rs)/2;
		if (c[k].a[mid].x>=vl) ans=min(ans,mid),rs=mid-1;
		else ls=mid+1; 
	}
	if (c[k].a[ans].x==vl&&ans!=n+1) 
	 x=min(x,c[k].a[ans].pos),y=max(y,c[k].a[ans].pos);
	ls=1; rs=r[k]-l[k]+1;  ans=0;
	while (ls<=rs) {
		int mid=(ls+rs)/2;
		if (c[k].a[mid].x<=vl) ans=max(ans,mid),ls=mid+1;
		else rs=mid-1;
	}
	if (c[k].a[ans].x==vl&&ans!=0) 
	  x=min(x,c[k].a[ans].pos),y=max(y,c[k].a[ans].pos);
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%I64d",&val[i]);
	int block=sqrt(n);
	int t=ceil(n*1.0/block);
	for (int i=1;i<=n;i++) belong[i]=(i-1)/block+1;
    for (int i=1;i<=t;i++) l[i]=n,r[i]=0;
    for (int i=1;i<=n;i++) {
    	int k=belong[i];
    	l[k]=min(i,l[k]);
    	r[k]=max(r[k],i);
	}
	for (int i=1;i<=t;i++) build(i);
	for (int i=1;i<=m;i++) {
		int opt,x,y; LL k;
		scanf("%d",&opt);
		if (opt==1) {
			scanf("%d%d%I64d",&x,&y,&k);
			if (belong[x]==belong[y]) {
				for (int i=x;i<=y;i++) val[i]+=k;
				build(belong[x]); continue;
			}
			for (int i=x;i<=r[belong[x]];i++) val[i]+=k;
			build(belong[x]);
			for (int i=l[belong[y]];i<=y;i++) val[i]+=k;
			build(belong[y]);
			for (int i=belong[x]+1;i<=belong[y]-1;i++) delta[i]+=k;
		}
		else {
			scanf("%I64d",&k);
			int x=n+1; int y=0;
			for (int i=1;i<=t;i++) 
			 calc(x,y,i,k);
			printf("%d\n",max(-1,y-x));
		}
	}
}