Codeforces Round #307 (Div. 2) 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 a_j = a_i = 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 a_i 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, ... a_n (1 ≤ a_i ≤ 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
【分析】给出一个数组a，然后q次操作，1 l r x表示将区间[l,r]所有数+1 ； 2 y表示询问这个数组中最左边的y和最右边的y的下标差为多少。
 可分块做。将所有块里面的数排序，（便于查找要查询的y）。对于更新操作，单独处理左端点和右端点的块，改变a[i];然后对于中间跨过的块，用lazy数组标记add值。
 对于查询操作，依次遍历每个块，二分查找y的位置即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 1000000000
#define met(a,b) memset(a,b,sizeof a)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
using namespace std;
const int N = 5e5+5;
const int M = 7e2+50;
int n,m;
int l[N],r[N],belong[N];
int cnt,num,x,v,ans;
int a[N],tot[M],lazy[M];
pair<int,int>p[M][M];
void init(){
    num=sqrt(n);
    cnt=n/num;
    if(n%num)cnt++;
    for(int i=1;i<=n;i++){
        belong[i]=(i-1)/num+1;
    }
    for(int i=1;i<=cnt;i++){
        l[i]=(i-1)*num+1;
        r[i]=min(n,i*num);
        for(int j=l[i];j<=r[i];j++){
            p[i][++tot[i]]=make_pair(a[j],j);
        }
        sort(p[i]+1,p[i]+1+tot[i]);
    }
}
void update1(int b,int x){
    if(lazy[b]+x>inf)lazy[b]=inf+1;
    else lazy[b]+=x;
}
void update2(int b,int ll,int rr,int x){
    if(ll==l[b]&&rr==r[b]){
        update1(b,x);
        return;
    }
    for(int i=1;i<=tot[b];i++){
        int po=p[b][i].second;
        if(po>=ll&&po<=rr){
            p[b][i].first+=x;
            if(p[b][i].first>inf)p[b][i].first=inf+1;
        }
    }
    sort(p[b]+1,p[b]+tot[b]+1);
}
int main() {
    int op,ll,rr,x,y;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    init();
    while(m--){
        scanf("%d",&op);
        if(op==1){
            scanf("%d%d%d",&ll,&rr,&x);
            if(belong[ll]==belong[rr])update2(belong[ll],ll,rr,x);
            else {
                for(int i=belong[ll]+1;i<=belong[rr]-1;i++)update1(i,x);
                update2(belong[ll],ll,r[belong[ll]],x);
                update2(belong[rr],l[belong[rr]],rr,x);
            }
        }
        else {
            scanf("%d",&y);
            int last=0,fir=0;
            for(int i=cnt;i>=1;i--){
                if(!last){
                    int po=upper_bound(p[i]+1,p[i]+tot[i]+1,make_pair(y-lazy[i],1000000))-p[i]-1;
                    if(p[i][po].first+lazy[i]==y){
                        fir=last=p[i][po].second;
                    }
                }
                int po=lower_bound(p[i]+1,p[i]+tot[i]+1,make_pair(y-lazy[i],0))-p[i];
                if(p[i][po].first+lazy[i]==y){
                    fir=p[i][po].second;
                }
            }
            if(!last)puts("-1");
            else printf("%d
",last-fir);
        }
    }
    return 0;
}