A - Zero Array （SDUT 2018 Autumn Individual Contest - I）

滴答滴答---题目链接滴答滴答--题目链接

You are given an array a consisting of n elements, and q queries. There are two types of queries, as follow:

• "1 p v" – An update query asks to change the value at position p in array a to v.
• "2" – A query asks to print the minimum number of required operations to convert array a to a zero array.

A zero array is defined as an array which all its elements are zeros. There is only one allowed operation to convert an array to a zero array. At each operation, you can choose a value x and subtract it from all non-zero elements in the array, such that no element will be negative after the operation.

Input

The first line contains an integer T (1 ≤ T ≤ 100), in which T is the number of test cases.

The first line of each test case consists of two integers n and q (1 ≤ n, q ≤ 105), in which n is the size of the array a, and q is the number of queries.

Then a line follows containing n elements a1, a2, ..., an (0 ≤ ai ≤ 109), giving the array a.

Then q lines follow, each line containing a query in the format described in the problem statement. It is guaranteed that the following constraints hold for the first type of queries: 1 ≤ p ≤ n, 0 ≤ v ≤ 109.

The sum of n and q overall test cases does not exceed 106 for each.

Output

For each query of the second type, print the minimum number of required operations to convert array a to a zero array. The queries must be answered in the order given in the input.

Example

Input

1
5 5
3 2 1 5 4
1 2 3
2
1 3 2
1 4 1
2

Output

4
4

#include
using namespace std;
int a[100001];
mapmp;
int main()
{
    int t,n,m,x,y,k;
    scanf("%d",&t);
    while(t--)
    {
        mp.clear();
        int ans=0;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]&&!mp[a[i]])ans++;
            mp[a[i]]++;
        }
        while(m--)
        {
            scanf("%d",&k);
            if(k==1)
            {
                scanf("%d%d",&x,&y);
                if(a[x]&&mp[a[x]]==1)ans--;
                mp[a[x]]--;
                if(y&&!mp[y])ans++;
                mp[y]++;
                a[x]=y;
            }
            else printf("%d\n",ans);
        }
    }
    return 0;
}