// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

struct edge {

    int to;     // 辺の行き先

    int weight; // 辺の重み

    edge(int t, int w) : to(t), weight(w) { }

};

using Graph = vector<vector<int>>;

using dou =long double;

string yes="yes";

string Yes="Yes";

string YES="YES";

string no="no";

string No="No";

string NO="NO";



template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

#define rep(i, n)         for(int i = 0; i < (int)(n); i++)

#define brep(n)           for(int bit=0;bit<(1<<n);bit++)

#define erep(i,container) for (auto i : container)

#define irep(i, n)        for(int i = n-1; i >= (int)0; i--)

#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)

#define reprep(i,j,h,w) rep(i,h)rep(j,w)

#define all(x) (x).begin(),(x).end()

#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];

#define pb push_back

#define pf push_front

#define lb lower_bound

#define ub upper_bound

#define fi first

#define se second

#define itn int

#define mp make_pair

#define sum(a) accumulate(all(a),0ll)

#define keta fixed<<setprecision

#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))

#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));

#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));

#define case std::cout << "Case #" <<qqq+1<<": "

#define res resize

#define as assign

#define ffor for(;;)

#define ppri(a,b) std::cout << a<<" "<<b << std::endl

#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl

#define aall(x,n) (x).begin(),(x).begin()+(n)

 

typedef long long ll;

typedef pair<int,int> P;

//typedef long long T;

const int INF = 2000000000;

const ll INF64 = 922337203685477580ll;

const ll MOD = 1000000007ll;

std::vector<int> al(300000,0);

void dfs(std::vector<std::vector<int>> &g,std::vector<ll> &v,int p){

    al[p]=1;

    erep(i,g[p]){

        if(al[i])continue;

        v[i]+=v[p];

        dfs(g,v,i);

    }

}

int main(){

    int n,q;

    std::cin >> n>>q;

    Graph g(n);

    rep(i,n-1){

        int a,b;

        std::cin >> a>>b;

        a--;b--;

        g[a].pb(b);

        g[b].pb(a);

    }

    std::vector<ll> v(n,0);

    rep(i,q){

        int p,x;

        std::cin >> p>>x;

        p--;

        v[p]+=x;

    }

//    ppri(g.size(),v.size());

    dfs(g,v,0);

  //  std::cout << 1 << std::endl;

    erep(i,v){

        std::cout << i <<  ' ';

    }

    std::cout  << std::endl;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

//#include <boost/multiprecision/cpp_int.hpp>

using namespace std;



using dou =long double;

string yes="yes";

string Yes="Yes";

string YES="YES";

string no="no";

string No="No";

string NO="NO";



template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

typedef long long ll;

typedef pair<int,int> P;

typedef pair<ll,ll> PL;

const ll mod = 1000000007ll;

//const ll mod = 1000003;

struct mint {

  ll x; // typedef long long ll;

  mint(ll x=0):x((x%mod+mod)%mod){}

  mint operator-() const { return mint(-x);}

  mint& operator+=(const mint a) {

    if ((x += a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator-=(const mint a) {

    if ((x += mod-a.x) >= mod) x -= mod;

    return *this;

  }

  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}

  mint operator+(const mint a) const { return mint(*this) += a;}

  mint operator-(const mint a) const { return mint(*this) -= a;}

  mint operator*(const mint a) const { return mint(*this) *= a;}

  mint pow(ll t) const {

    if (!t) return 1;

    mint a = pow(t>>1);

    a *= a;

    if (t&1) a *= *this;

    return a;

  }



  // for prime mod

  mint inv() const { return pow(mod-2);}

  mint& operator/=(const mint a) { return *this *= a.inv();}

  mint operator/(const mint a) const { return mint(*this) /= a;}

};

istream& operator>>(istream& is, const mint& a) { return is >> a.x;}

ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}









#define rep(i, n)         for(ll i = 0; i < (ll)(n); i++)

#define brep(n)           for(int bit=0;bit<(1<<n);bit++)

#define bbrep(n)           for(int bbit=0;bbit<(1<<n);bbit++)

#define erep(i,container) for (auto &i : container)

#define itrep(i,container) for (auto i : container)

#define irep(i, n)        for(ll i = n-1; i >= (ll)0ll; i--)

#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)

#define reprep(i,j,h,w) rep(i,h)rep(j,w)

#define all(x) (x).begin(),(x).end()

#define rall(x) (x).rbegin(),(x).rend()

#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];

#define pb push_back

#define pf push_front

#define query int qq;std::cin >> qq;rep(qqq,qq)

#define lb lower_bound

#define ub upper_bound

#define fi first

#define se second

#define itn int

#define mp make_pair

#define sum(a) accumulate(all(a),0ll)

#define keta fixed<<setprecision

#define vout(a) erep(qxqxqx,a)std::cout << qxqxqx << ' ';std::cout  << std::endl;

#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))

//#define vvector(name,typ,m,n)vector<vector<typ> > name(m,vector<typ> (n))

#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));

#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));

#define case std::cout <<"Case #" <<qqq+1<<":"

#define res resize

#define as assign

#define ffor for(;;)

#define ppri(a,b) std::cout << a<<" "<<b << std::endl

#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl

#define aall(x,n) (x).begin(),(x).begin()+(n)

#define ssum(a) accumulate(a,0ll) 

#define stirng string

//#define grid_input(a,type) int h,w;std::cin >> h>>w;vvector(a,type,h,w,0);reprep(i,j,h,w)std::cin >> a[i][j];



//typedef long long T;

ll ceil(ll a,ll b){

    return ((a+b-1)/b);

}

const int INF = 2000000000;

//const ll INF64 =3223372036854775807ll;

//const ll INF64 = 9223372036854775807ll;

const ll INF64 = 9223372036854775ll;

const ll MOD = 1000000007ll;

//const ll MOD = 1000003ll;

const ll OD = 1000000000000007ll;

const dou pi=3.141592653589793;

long long modpow(long long a, long long n) { //累乗の余剰

    long long res = 1;

    while (n > 0) {

        if (n & 1) res = res * a % MOD;

        a = a * a % MOD;

        n >>= 1;

    }

    return res;

    

}



//メモ

//tuple<hoge,huga,foo> tのget関数はget<i>(t)

//resizeはメンバ関数

//三項演算子を覚えよう

// a?b:cと書くと aならばbの処理をする

//i本目にbitをたてる　-> (1<<i)

int main(){

    int n,qq;

    std::cin >> n>>qq;

    std::vector<std::vector<int>> g(n);

    rep(i,n-1){

        int a,b;

        std::cin >> a>>b;

        a--;b--;

        g[a].pb(b);

        g[b].pb(a);

    }

    std::vector<int> c(n);

    rep(i,qq){

        itn p,x;

        std::cin >> p>>x;

        p--;

        c[p]+=x;

    }

    std::vector<int> ans(n,-1);

    ans[0]=c[0];

    std::queue<int> q;

    q.push(0);

    while(!q.empty()){

        int p=q.front();

        erep(i,g[p]){

            if(ans[i]!=-1)continue;

            q.push(i);

            ans[i]=ans[p]+c[i];

        }

        q.pop();

    }

    vout(ans);

    

}

// Fastest known solution for this problem: