[Codefroces 1230E] Kamil and Making a Stream(List dfs gcd)

题意是求树中所有直链的gcd的总和。

做法是dfs把前缀的所有信息传下去，暴力求值。

为什么可以暴力呢？x = gcd(x,a);x会怎么变呢？要么不变，x，a的最大公因数就是x，要么x变为x的一个因子，也就是说最大是x/2，不出log次计算，x就会变成1

这也就是说维护每个点到根的后缀gcd，最多只有log个不同的数，用链表存起来就可以。

#include <iostream>
#include <queue>
#include <cstring>
#include <list>
#include <algorithm>using namespace std;#define MAXX 201000
typedef long long LL;struct E {int to,next;
}g[MAXX];
int front[MAXX],tot;void Add(int from,int to) {g[++tot].to = to;g[tot].next = front[from];front[from] = tot;
}LL gcd(LL a,LL b) {return b==0 ? a:gcd(b,a%b);
}LL v[MAXX],ans;
const LL Mod = 1e9+7;
typedef list<pair<LL,int> > List;
#define val first
#define num second
void dfs(int t,int fa,List st) {
/*  if(!st.empty()) {List::reverse_iterator it = st.rbegin();  (*it).val = gcd((*it).val,v[t]);for (it++; it != st.rend(); it++) (*it).val = gcd(v[t],(*it).val);st.push_back({v[t],1});} else st.push_back({v[t],1});List::iterator tp = st.begin(),tpp = tp,del;for (tp++; tp != st.end(); )if ((*tp).val == (*tpp).val) {(*tpp).num += (*tp).num;del = tp;tp++;st.erase(del);} else tp++;*/List stt;stt.emplace_back(v[t],1);for (auto it : st) {LL n_va = gcd(it.val,v[t]);if (n_va == stt.back().val) stt.back().num += it.num;else stt.emplace_back(n_va,it.num);}  st.swap(stt);for (auto it : st)ans = (ans+it.val*it.num%Mod)%Mod;for (int i = front[t]; i; i = g[i].next) {int to = g[i].to;if (to != fa) dfs(to,t,st);}st.clear();
} int main() {int n; cin >> n;for (int i = 1; i <= n; i++) scanf("%I64d",&v[i]);for (int i = 1; i < n; i++) {int u,v;scanf("%d%d",&u,&v);Add(u,v);Add(v,u);}List tp;dfs(1,-1,tp);printf("%I64d",ans);return 0;
}

Kamil likes streaming the competitive programming videos. His MeTube channel has recently reached 100100 million subscribers. In order to celebrate this, he posted a video with an interesting problem he couldn't solve yet. Can you help him?

You're given a tree — a connected undirected graph consisting of nn vertices connected by n−1n−1 edges. The tree is rooted at vertex 11. A vertex uu is called an ancestor of vv if it lies on the shortest path between the root and vv. In particular, a vertex is an ancestor of itself.

Each vertex vv is assigned its beauty xvxv — a non-negative integer not larger than 10121012. This allows us to define the beauty of a path. Let uu be an ancestor of vv. Then we define the beauty f(u,v)f(u,v) as the greatest common divisor of the beauties of all vertices on the shortest path between uu and vv. Formally, if u=t1,t2,t3,…,tk=vu=t1,t2,t3,…,tk=v are the vertices on the shortest path between uu and vv, then f(u,v)=gcd(xt1,xt2,…,xtk)f(u,v)=gcd(xt1,xt2,…,xtk). Here, gcdgcd denotes the greatest common divisor of a set of numbers. In particular, f(u,u)=gcd(xu)=xuf(u,u)=gcd(xu)=xu.

Your task is to find the sum

∑u is an ancestor of vf(u,v).∑u is an ancestor of vf(u,v).

As the result might be too large, please output it modulo 109+7109+7.

Note that for each yy, gcd(0,y)=gcd(y,0)=ygcd(0,y)=gcd(y,0)=y. In particular, gcd(0,0)=0gcd(0,0)=0.

Input

The first line contains a single integer nn (2≤n≤1000002≤n≤100000) — the number of vertices in the tree.

The following line contains nn integers x1,x2,…,xnx1,x2,…,xn (0≤xi≤10120≤xi≤1012). The value xvxv denotes the beauty of vertex vv.

The following n−1n−1 lines describe the edges of the tree. Each of them contains two integers a,ba,b (1≤a,b≤n1≤a,b≤n, a≠ba≠b) — the vertices connected by a single edge.

Output

Output the sum of the beauties on all paths (u,v)(u,v) such that uu is ancestor of vv. This sum should be printed modulo 109+7109+7.

Examples

input

Copy
5
4 5 6 0 8
1 2
1 3
1 4
4 5
output

Copy
42
input

Copy
7
0 2 3 0 0 0 0
1 2
1 3
2 4
2 5
3 6
3 7
output

Copy
30
Note

The following figure shows all 1010 possible paths for which one endpoint is an ancestor of another endpoint. The sum of beauties of all these paths is equal to 4242:

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