一棵n个点的树，每个点的初始权值为1。对于这棵树有q个操作，每个操作为以下四种操作之一：
+ u v c：将u到v的路径上的点的权值都加上自然数c；
- u1 v1 u2 v2：将树中原有的边(u1,v1)删除，加入一条新边(u2,v2)，保证操作完之后仍然是一棵树；
* u v c：将u到v的路径上的点的权值都乘上自然数c；
/ u v：询问u到v的路径上的点的权值和，求出答案对于51061的余数。

第一行两个整数n，q
接下来n-1行每行两个正整数u，v，描述这棵树
接下来q行，每行描述一个操作

对于每个/对应的答案输出一行

3 2
1 2
2 3
* 1 3 4
/ 1 1

4

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int mod=51061;
const int maxn=50010;int fir[maxn],nxt[maxn*2],to[maxn*2],cnt;void addedge(int a,int b){nxt[++cnt]=fir[a];fir[a]=cnt;to[cnt]=b;
}int fa[maxn],rt[maxn];
int ch[maxn][2],key[maxn],sz[maxn];
int sum[maxn],flip[maxn],add[maxn],mul[maxn];void DFS(int x){rt[x]=sz[x]=key[x]=sum[x]=mul[x]=1;for(int i=fir[x];i;i=nxt[i])    if(to[i]!=1&&!fa[to[i]]){fa[to[i]]=x;DFS(to[i]);}
}void Add(int x,int d){if(!x)return;key[x]=(key[x]+d)%mod;add[x]=(add[x]+d)%mod;sum[x]=(sum[x]+sz[x]*d%mod)%mod;
}void Mul(int x,int k){key[x]=(key[x]*k)%mod;sum[x]=(sum[x]*k)%mod;add[x]=(add[x]*k)%mod;
}void Flip(int x){swap(ch[x][0],ch[x][1]);flip[x]^=1;
}void Push_down(int x){if(flip[x]){Flip(ch[x][0]);Flip(ch[x][1]);flip[x]=0;}if(mul[x]!=1){Mul(ch[x][0],mul[x]);Mul(ch[x][1],mul[x]);mul[x]=1;}if(add[x]){Add(ch[x][0],add[x]);Add(ch[x][1],add[x]);add[x]=0;}
}void Push_up(int x){sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1;sum[x]=(sum[ch[x][0]]+sum[ch[x][1]]+key[x])%mod;
}void Rotate(int x){int y=fa[x],g=fa[y],c=ch[y][1]==x;ch[y][c]=ch[x][c^1];ch[x][c^1]=y;fa[ch[y][c]]=y;fa[y]=x;fa[x]=g;if(rt[y])rt[x]=1,rt[y]=0;else ch[g][ch[g][1]==y]=x;Push_up(y);
}void P(int x){if(!rt[x])P(fa[x]);Push_down(x);
}void Splay(int x){P(x);for(int y=fa[x];!rt[x];Rotate(x),y=fa[x])if(!rt[y])Rotate((ch[fa[y]][1]==y)==(ch[y][1]==x)?y:x);Push_up(x);
}void Access(int x){int y=0;while(x){Splay(x);rt[ch[x][1]]=1;rt[ch[x][1]=y]=0;Push_up(x);x=fa[y=x];}
}void Make_RT(int x){Access(x);Splay(x);Flip(x);
}void Link(int x,int y){Make_RT(x);fa[x]=y;
}void Cut(int x,int y){Make_RT(x);Splay(y);fa[ch[y][0]]=fa[y];fa[y]=0;rt[ch[y][0]]=1;ch[y][0]=0;Push_up(y);
}void Lca(int &x,int &y){Access(y);y=0;while(true){Splay(x);if(!fa[x])break;rt[ch[x][1]]=1;rt[ch[x][1]=y]=0;Push_up(x);x=fa[y=x];}
}void ADD(int x,int y,int d){Lca(x,y);Add(y,d);Add(ch[x][1],d);key[x]=(key[x]+d)%mod;Push_up(x);
}void MUL(int x,int y,int k){Lca(x,y);Mul(y,k);Mul(ch[x][1],k);key[x]=(key[x]*k)%mod;Push_up(x);
}int Query(int x,int y){Lca(x,y);int ret=(key[x]+sum[y]+sum[ch[x][1]])%mod;return ret;
}int n,Q;
char op[5];
int main(){
#ifndef ONLINE_JUDGEfreopen("nt2012_wym_tree.in","r",stdin);freopen("nt2012_wym_tree.out","w",stdout);
#endifscanf("%d%d",&n,&Q);for(int i=1,a,b;i<n;i++){scanf("%d%d",&a,&b);addedge(a,b);addedge(b,a);}DFS(1);int x,y,c,u,v;while(Q--){scanf("%s",op);if(op[0]=='-'){scanf("%d%d",&u,&v);Cut(u,v);scanf("%d%d",&u,&v);Link(u,v);}else if(op[0]=='/'){scanf("%d%d",&x,&y);printf("%d\n",Query(x,y));}else{scanf("%d%d%d",&x,&y,&c);if(op[0]=='+')ADD(x,y,c);elseMUL(x,y,c);    }}return 0;
}