1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define R register int
  4 #define AC 101000
  5 #define ac 500000
  6 #define maxn 2010000
  7 #define p 1000000007
  8 #define LL long long
  9
 10 int n, m, top, cnt, w;
 11 LL rnt = 1;
 12 int Head[maxn], Next[maxn], date[maxn], tot;
 13 int father[AC], up[AC];
 14 LL inv[AC], Size[AC], ans[AC];
 15 struct line{
 16     int x, y;
 17     friend bool operator < (line a, line b)
 18     {
 19         if(a.x != b.x) return a.x < b.x;
 20         else return a.y < b.y;
 21     }
 22 }road[AC];
 23
 24 struct node{
 25     int fa, x;
 26 }s[AC];
 27
 28 map<line, int> MAP;
 29
 30 inline int read()
 31 {
 32     int x = 0;char c = getchar();
 33     while(c > '9' || c < '0') c = getchar();
 34     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
 35     return x;
 36 }
 37
 38 inline void add(int f, int w){
 39     date[++tot] = w, Next[tot] = Head[f], Head[f] = tot;
 40 }
 41
 42 inline int find(int x){
 43     while(father[x] != x) x = father[x];
 44     return x;
 45 }
 46
 47 inline void link(int &ret, int x, int y)
 48 {
 49     int fx = find(x), fy = find(y);
 50     if(fx == fy) return ;
 51     ++ ret;//只有连接了才要加ret
 52     if(Size[fx] > Size[fy]) swap(fx, fy);
 53     rnt = rnt * inv[Size[fx]] % p * inv[Size[fy]] % p;
 54     father[fx] = fy, Size[fy] += Size[fx];
 55     s[++top] = (node){fy, fx};
 56     rnt = rnt * Size[fy] % p;
 57 }
 58
 59 inline void cut()
 60 {
 61     node x = s[top --];//撤销
 62     rnt = rnt * inv[Size[x.fa]] % p;
 63     father[x.x] = x.x, Size[x.fa] -= Size[x.x];//断开连接
 64     rnt = rnt * Size[x.fa] % p * Size[x.x] % p;//注意删除一个点之后要把父亲修改为自己，而不是0
 65 }
 66
 67 void solve(int x, int l, int r)//当前区间编号，区间范围
 68 {
 69     int now, ret = 0;
 70     for(R i = Head[x]; i ; i = Next[i])
 71     {
 72         now = date[i];
 73         link(ret, road[now].x, road[now].y);
 74     }
 75     int mid = (l + r) >> 1;
 76     if(l == r) ans[l] = rnt;
 77     if(l != r) solve(x * 2, l, mid), solve(x * 2 + 1, mid + 1, r);
 78     for(R i = 1; i <= ret; i ++) cut();
 79 }
 80
 81 void pre()
 82 {
 83     n = read(), m = read();
 84     inv[0] = inv[1] = 1;
 85     for(R i = 1; i <= n; i ++) father[i] = i, Size[i] = 1;
 86     for(R i = 2; i <= n; i ++)
 87         inv[i] = (p - p / i) * inv[p % i] % p;
 88 }
 89
 90 void change(int x, int l, int r, int ll, int rr)
 91 {
 92     if(l == ll && r == rr){add(x, w); return ;}
 93     int mid = (l + r) >> 1;
 94     if(rr <= mid) change(x * 2, l, mid, ll, rr);
 95     else if(ll > mid) change(x * 2 + 1, mid + 1, r, ll, rr);
 96     else change(x * 2, l, mid, ll, mid), change(x * 2 + 1, mid + 1, r, mid + 1, rr);
 97 }
 98
 99 void work()
100 {
101     int tmp;
102     for(R i = 1; i <= m; i ++)
103     {
104         int opt = read(), a = read(), b = read();
105         if(a > b) swap(a, b);
106         if(!MAP[(line){a, b}])
107             MAP[(line){a, b}] = ++ cnt, tmp = cnt, road[cnt] = (line){a, b};
108         else tmp = MAP[(line){a, b}];//获取编号
109         if(opt == 1) up[tmp] = i;//存下这条边的出现时间
110         else w = tmp, change(1, 1, m, up[tmp], i - 1), up[tmp] = 0;
111     }
112     for(R i = 1; i <= cnt; i ++)
113         if(up[i]) w = i, change(1, 1, m, up[i], m);
114     solve(1, 1, m);
115     for(R i = 1; i <= m; i ++) printf("%lld\n", ans[i]);
116 }
117
118 int main()
119 {
120     //freopen("in.in", "r", stdin);
121     pre();
122     work();
123     //fclose(stdin);
124     return 0;
125 }