有多组样例，每组样例第一行输入两个正整数n,m(2 <= n<=50000，1<=m <= 50000),接下来n-1行，每行3个正整数a b c，(1 <= a,b <= n , a != b , 1 <= c <= 1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行，表示m个操作，操作有两种：一. 0 a b，表示更新第a条路的过路费为b，1 <= a <= n-1 ； 二. 1 a b ， 表示询问a到b最少要花多少过路费。

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<algorithm>

using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")

#define ls i<<1

#define rs ls | 1

#define mid ((ll+rr)>>1)

#define pii pair<int,int>

#define MP make_pair

typedef long long LL;

const long long INF = 1e18;

const double Pi = acos(-1.0);

const int N = 2e5+, M = 5e5+, inf = 2e9, mod = ;

struct edge{int to,next,value;}e[N * ];

struct Lin {int u; int v; int w;

    Lin(int u = , int v = , int w = ) : u(u), v(v), w(w) {}

}L[N*];

LL sum[N];

int head[N],t=,f[N],q,n,top[N],siz[N],son[N],pos[N],val[N],deep[N],tot;

void init() {

    t = ;

    memset(head,,sizeof(head));

    deep[] = ;

    siz[] = ;

    f[] = ;

    tot = ;

}

void add(int u,int v)

{e[t].next=head[u];e[t].to=v;head[u]=t++;}

void dfs1(int u,int fa) {

        siz[u] = ;

        son[u] = ;

        f[u] = fa;

        deep[u] = deep[fa] + ;

        for(int i = head[u]; i; i = e[i].next) {

            int to = e[i].to;

            if(to == fa) continue;

            dfs1(to,u);

            siz[u] += siz[to];

            if(siz[to] > siz[son[u]]) son[u] = to;

        }

}

void dfs2(int u,int chan) {

        top[u] = son[chan]==u?top[chan]:u;

        pos[u] = ++tot;

        if(son[u]) dfs2(son[u],u);

        for(int i = head[u]; i; i = e[i].next) {

            int to = e[i].to;

            if(to == son[u] || to == chan) continue;

            dfs2(to,u);

        }

}

void build(int i,int ll,int rr)

{

    sum[i] = ;

    if(ll == rr) {

        sum[i] = val[ll];

        return ;

    }

    build(ls,ll,mid), build(rs,mid+,rr);

    sum[i] = sum[ls] + sum[rs];

}

void update(int i,int ll,int rr,int x,int c) {

    if(x == ll && x == rr) {

        sum[i] = c;

        return ;

    }

    if(x <= mid) update(ls,ll,mid,x,c);

    else update(rs,mid+,rr,x,c);

    sum[i] = sum[ls] + sum[rs];

}

LL query(int i,int ll,int rr,int x,int y) {

    if(x == ll && rr == y) {

        return sum[i];

    }

    if(y <= mid) return query(ls,ll,mid,x,y);

    else if(x > mid) return query(rs,mid+,rr,x,y);

    else return query(ls,ll,mid,x,mid) + query(rs,mid+,rr,mid+,y);

}

LL sub_query(int x,int y,LL ret = ) {

        while (top[x] != top[y])

        {

            if(deep[top[x]] < deep[top[y]]) swap(x,y);

            ret += query(,,n,pos[top[x]],pos[x]);

            x = f[top[x]];

        }

        if(x == y) return ret;

        if(deep[x] > deep[y]) swap(x,y);

        return ret + query(,,n,pos[x]+, pos[y]);

}

void solve() {

        init();

        for(int i = ; i <= n-; ++i) {

                int a,b,c;

                scanf("%d%d%d",&a,&b,&c);

                L[i] = Lin(a,b,c);

                add(a,b);add(b,a);

        }

        dfs1(,);

        dfs2(,);

        val[] = ;

        for(int i = ; i <= n-; ++i) {

            if(deep[L[i].u] < deep[L[i].v]) swap(L[i].u, L[i].v);

            val[pos[L[i].u]] = L[i].w;

        }//cout<<1<<endl;

        build(,,n);

        while(q--){

            int op,x,y;

            scanf("%d%d%d",&op,&x,&y);

            if(op == )

                update(,,n,pos[L[x].u],y);

            else printf("%I64d\n",sub_query(x,y));

        }

}

int main() {

        while(~scanf("%d%d",&n,&q)) {solve();}return ;

}