hdu_1011

// hdu 1011
// tree dp
// Feb.16 2015

#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>

#define MAXN 111

std::vector<int> tunnel[MAXN];
int n, m, dp[MAXN][MAXN], bugs[MAXN], possi[MAXN];
// dp(num)(i) is the total possibility assign (i) Troopers to capture
// the tree whose root is (num)
bool vis[MAXN];

void dfs(int num)
{
    
    vis[num] = true;

    // for the cave numbered (num)
    // at least ( ceil(bugs[num]/20) ) Troopers are need 
    for(int i = bugs[num]; i <= m; ++i)
        dp[num][i] = possi[num];

    for(int i = 0; i < tunnel[num].size(); ++i){

        if(vis[ tunnel[num][i] ] == true)
            continue;

        dfs( tunnel[num][i] );
        // 对于一个根节点编号为num的子树
        // j为num节点的指派Trooper数量 
        // k为除num节点以外其他节点上Trooper数量
        for(int j = m; j >= bugs[num]; --j)
            for(int k = 1; k + bugs[num] <= j; ++k)
                dp[num][j] = std::max( dp[num][bugs[num]], dp[num][bugs[num]] + dp[ tunnel[num][i] ][k] );

    }
}

int main(int argc, char const *argv[])
{
    // freopen("in", "r", stdin);
    while(scanf("%d%d", &n, &m)){
        if(n == -1 && m == -1)
            return 0;
        memset(dp, 0, sizeof(dp));
        memset(vis, false, sizeof(vis));
        for(int i = 1; i <= n; ++i){
            scanf("%d%d", bugs+i, possi + i);
            bugs[i] = ceil(double(bugs[i])/20);
            // transform how many bugs into how many Troopers can defeat bugs in the cave
            tunnel[i].clear();
        }
        for(int i = 1, a, b; i < n; ++i){
            scanf("%d%d", &a, &b);
            tunnel[a].push_back(b);
            tunnel[b].push_back(a);
        }

        // if there is 0 bug in a cave 
        // we still need at least 1 Trooper to capture this cave
        if( m == 0 ){
            printf("0\n");
            continue;
        }
        
        dfs(1);
        // the aim status
        printf("%d\n", dp[1][m]);
    }
    return 0;
}