There is a skyscraping tree standing on the playground of Nanjing University of Science and Technology. On each branch of the tree is an integer (The tree can be treated as a connected graph with N vertices, while each branch can be treated as a vertex). Today the students under the tree are considering a problem: Can we find such a chain on the tree so that the multiplication of all integers on the chain (mod 10⁶ + 3) equals to K?
Can you help them in solving this problem?

 

 

There are several test cases, please process till EOF.
Each test case starts with a line containing two integers N(1 <= N <= 10⁵) and K(0 <=K < 10⁶ + 3). The following line contains n numbers v_i(1 <= v_i < 10⁶ + 3), where vi indicates the integer on vertex i. Then follows N - 1 lines. Each line contains two integers x and y, representing an undirected edge between vertex x and vertex y.

 

 

For each test case, print a single line containing two integers a and b (where a < b), representing the two endpoints of the chain. If multiply solutions exist, please print the lexicographically smallest one. In case no solution exists, print “No solution”(without quotes) instead.
For more information, please refer to the Sample Output below.

 

 

 

 

 

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<vector>

#include<map>

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 = 1e5+, M = 1e6+, inf = 2e9, mod = ;

int head[N],vis[N],f[N],siz[N],id[N],n,t = ,ansl,ansr,allnode,root;

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

LL mp[M],inv[M],v[M],K,deep[M];

void add(int u,int v) {e[t].next=head[u];e[t].to=v;head[u]=t++;}

void getroot(int u,int fa) {

        f[u] = ;

        siz[u] =  ;

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

            int to = e[i].to;

            if(vis[to] || to == fa) continue;

            getroot(to,u);

            siz[u] += siz[to];

            f[u] = max(f[u],siz[to]);

        }

        f[u] = max(f[u], allnode - siz[u]);

        if(f[u] < f[root]) root = u;

}

void getdeep(int u,int fa,LL now) {

    deep[++deep[]] = now*v[u]%mod;

    id[deep[]] = u;

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

        int to = e[i].to;

        if(vis[to] || to == fa) continue;

        getdeep(to,u,now*v[u]%mod);

    }

}

void update(int u,int x,int y) {

        int tmp = mp[inv[x*v[u]%mod]*K%mod];

        if(!tmp) return ;

        if(y > tmp) swap(y,tmp);

        if(y < ansl || (y == ansl && tmp < ansr)) ansl = y, ansr = tmp;

}

void work(int u){

        vis[u] = ;

        mp[] = u;

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

            int to = e[i].to;

            if(vis[to]) continue;

            deep[] = ;

            getdeep(to,u,);

            for(int j = ; j <= deep[]; ++j) update(u,deep[j],id[j]);

            for(int j = ; j <= deep[]; ++j) if(!mp[deep[j]] || mp[deep[j]] > id[j])mp[deep[j]] = id[j];

        }

        mp[] = ;

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

            int to = e[i].to;

            if(vis[to]) continue;

            deep[] = ;

            getdeep(to,u,);

            for(int j = ; j <= deep[]; ++j)  mp[deep[j]] = ;

        }

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

            int to = e[i].to;

            if(vis[to]) continue;

            root = ;

            allnode = siz[to];

            getroot(e[i].to,root);

            work(root);

         }

}

int main() {

    inv[]=;

    for(int i=;i<mod;i++){int a=mod/i,b=mod%i;inv[i]=(inv[b]*(-a)%mod+mod)%mod;}

    while(~scanf("%d%I64d",&n,&K)) {

        t = ;memset(head,,sizeof(head));

        memset(vis,,sizeof(vis));

        ansl = ansr = inf;

        for(int i = ; i <= n; ++i) scanf("%I64d",&v[i]);

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

            int u,v;

            scanf("%d%d",&u,&v);

            add(u,v);

            add(v,u);

        }

        f[]=inf;

        allnode=n;root=;

        getroot(,);

        work(root);

        if(ansl == inf) puts("No solution");else

        printf("%d %d\n",ansl,ansr);

    }

    return ;

}