When you play the game of thrones, you win, or you die. There is no middle ground.

Cersei Lannister, A Game of Thrones by George R. R. Martin

There are n nobles, numbered from 1 to n. Noble i has a power of i. There are also m "friendships". A friendship between nobles a and b is always mutual.

A noble is defined to be vulnerable if both of the following conditions are satisfied: 

  * the noble has at least one friend, and 
  * all of that noble's friends have a higher power. 



You will have to process the following three types of queries. 

  1. Add a friendship between nobles u and v. 
  2. Remove a friendship between nobles u and v. 
  3. Calculate the answer to the following process. 



The process: all vulnerable nobles are simultaneously killed, and all their friendships end. Then, it is possible that new nobles become vulnerable. The process repeats itself until no nobles are vulnerable. It can be proven that the process will end in finite time. After the process is complete, you need to calculate the number of remaining nobles.

Note that the results of the process are not carried over between queries, that is, every process starts with all nobles being alive!

Input

The first line contains the integers n and m (1 ≤ n ≤ 2⋅ 10^5, 0 ≤ m ≤ 2⋅ 10^5) — the number of nobles and number of original friendships respectively.

The next m lines each contain the integers u and v (1 ≤ u,v ≤ n, u ≠ v), describing a friendship. No friendship is listed twice.

The next line contains the integer q (1 ≤ q ≤ 2⋅ {10}^{5}) — the number of queries. 

The next q lines contain the queries themselves, each query has one of the following three formats. 

  * 1 u v (1 ≤ u,v ≤ n, u ≠ v) — add a friendship between u and v. It is guaranteed that u and v are not friends at this moment. 
  * 2 u v (1 ≤ u,v ≤ n, u ≠ v) — remove a friendship between u and v. It is guaranteed that u and v are friends at this moment. 
  * 3 — print the answer to the process described in the statement. 

Output

For each type 3 query print one integer to a new line. It is guaranteed that there will be at least one type 3 query.

Examples

Input


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


Output


2
1


Input


4 3
2 3
3 4
4 1
1
3


Output


1

Note

Consider the first example. In the first type 3 query, we have the diagram below.

In the first round of the process, noble 1 is weaker than all of his friends (2 and 3), and is thus killed. No other noble is vulnerable in round 1. In round 2, noble 3 is weaker than his only friend, noble 4, and is therefore killed. At this point, the process ends, and the answer is 2.

<image>

In the second type 3 query, the only surviving noble is 4.

The second example consists of only one type 3 query. In the first round, two nobles are killed, and in the second round, one noble is killed. The final answer is 1, since only one noble survives.

<image>
3 0
2
1 2 1
3
 20 0
1
3
 1000 0
1
3
 20 0
20
3
3
1 17 19
1 19 20
1 4 8
1 9 18
1 1 14
1 16 19
1 10 19
1 1 7
1 6 10
1 18 19
1 9 16
1 15 17
1 11 12
1 10 15
3
1 2 9
1 1 13
1 16 20
 20 53
1 2
1 4
1 5
1 6
1 7
1 8
1 20
2 3
2 6
2 16
2 17
3 6
3 18
3 20
4 6
4 7
4 8
4 9
4 14
4 18
5 11
5 13
5 17
5 18
5 19
6 8
6 13
7 8
7 10
7 15
7 17
7 18
8 15
8 16
9 10
9 15
10 11
10 17
10 20
11 15
11 16
12 14
12 15
12 16
12 19
14 16
14 19
16 17
16 19
16 20
17 19
17 20
18 19
1
3

2
 20
 1000
 20
20
9
 4

// clang-format
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace chrono;
using namespace __gnu_pbds;
    
#ifndef shivashr
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#endif

/*---------------------------------------------------------------------------------*/    
#define int long long
typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef long double f80;
typedef pair<int, int> pii;
typedef gp_hash_table<int, int> umap;
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag,
        tree_order_statistics_node_update> oset;
typedef pair<pii, int> piii;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<pii> vii;
typedef vector<string> vs;
    
/*---------------------------------------------------------------------------------*/   

// #define INF 1000000000000000000ll
// #define MOD 2000000000000000000ll
// #define MOD 998244353
// #define EPS 1e-7
#define PI 3.1415926535897932385
// #define PI acos(-1)
    
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ff first
#define ss second
#define B begin()
#define E end()
#define all(ar) ar.begin(), ar.end()
#define sl(x) (int)x.length()
#define sz(x) (int)x.size();
    
#define fr(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
#define FOR(i, n) for (int i = 0, _n = (n); i < _n; i++)
#define repr(i, n) for (int i = n - 1; i >= 0; i--)
#define frr(i, a, b) for (int i = (a), _b = (b); i >= _b; i--)
#define foreach(it, ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fil(ar, val) memset(ar, val, sizeof(ar))  // 0x3f for inf, 0x80 for -INF can also use with pairs  

template<typename T, typename U> static inline void amin(T &x, U y) {if (y < x) x = y; }
template<typename T, typename U> static inline void amax(T &x, U y) { if (x < y) x = y; }

/*---------------------------------------------------------------------------------*/   

#ifdef shivashr
template<typename T>
void __p(T a) { cout << a << " "; }
template<typename T>
void __p(std::vector<T> a) { cout << "{ "; for (auto p : a) __p(p); cout << "}"; }
template<typename T, typename F>
void __p(pair<T, F> a) { cout << "{ "; __p(a.first); __p(a.second); cout << "}"; }
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) { __p(a1); __p(a...); }
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
    cout<<name<<" : ";__p(arg1);cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
    int bracket=0,i=0;
    for(; ;i++)
        if(names[i]==','&&bracket==0)
            break;
        else if(names[i]=='(')
            bracket++;
        else if(names[i]==')')
            bracket--;
    const char *comma=names+i;
    cout.write(names,comma-names)<<" : ";
    __p(arg1);
    cout<<"| ";
    __f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" "; __f(#__VA_ARGS__, __VA_ARGS__)
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
    
// mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
// inline bool equals(double a, double b) { return fabs(a - b) < 1e-9; }
// ll modpow(ll b, ll e, ll mod=MOD) {
//     ll ans=1; for(;e;b=b*b%mod,e/=2) if(e&1) ans=ans*b%mod; return ans; }
///////////////////////////////////////////////////////////////////////////////////////////////////////////
    
    
void solve(){
    int n, m;
    cin>>n>>m;
    map<int,int> mp;
    int u,v;
    int an=0;
    FOR(i,m){
        cin>>u>>v;
        mp[min(u,v)]++;
    }
    an = n-mp.size();
    int q;
    cin>>q;
    while(q--){
        int x;
        cin>>x;
        if(x==3){
            cout<<n-mp.size()<<endl;
        }
        else if(x==1){
            cin>>u>>v;
            if(mp[min(u,v)]==0){
                an--;
            }
            mp[min(u,v)]++;
        }
        else if(x==2){
            cin>>u>>v;
            mp[min(u,v)]--;
            if(mp[min(u,v)]==0){
                an++; 
            }
        }
    }
}
    
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(15);   cout<<fixed;
#ifdef shivashr
    cin.exceptions(cin.failbit);
    freopen("i.txt", "r", stdin);
    freopen("o.txt", "w", stdout);
#endif
    int t=1;
    // cin>>t;
    fr(i,1,t)
        solve();
   
    return 0;
}
