/*51 nod 1610 路径计数(Moblus+dp)problem:路径上所有边权的最大公约数定义为一条路径的值。给定一个有向无环图。T次修改操作，每次修改一条边的边权，每次修改后输出有向无环图上路径的值为1的路径数量(对1,000,000,007取模)。solve:感觉直接在图上求GCD的话很麻烦,而且还涉及到修改.后来发现可以考虑通过容斥来求GCD,这样的话就转换成了图上面长度为i的路径的个数.开始时记录路径长度w[i]以及它的约数. (w[i] = 4的话, 可以看成有 1,2,4三条边)于是通过枚举gcd便能够在 100*n*n内求出来所有路径值的情况.在修改的时候,可以发现只会 影响被移除的数和添加的数以及它们的约数. 处理一下然后通过moblus实现容斥就能求出gcd = 1的情况.hhh-2016/09/09-20:59:44*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <stdio.h>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define scanfd(a) scanf("%lf",&a)#define key_val ch[ch[root][1]][0]#define eps 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1000000007;const int maxn = 135;const double PI = acos(-1.0);const int limit = 33;template<class T> void read(T&num){    char CH;    bool F=false;    for(CH=getchar(); CH<‘0‘||CH>‘9‘; F= CH==‘-‘,CH=getchar());    for(num=0; CH>=‘0‘&&CH<=‘9‘; num=num*10+CH-‘0‘,CH=getchar());    F && (num=-num);}int stk[70], tp;template<class T> inline void print(T p){    if(!p)    {        puts("0");        return;    }    while(p) stk[++ tp] = p%10, p/=10;    while(tp) putchar(stk[tp--] + ‘0‘);    putchar(‘\n‘);}int n,m,q,id;ll dp[maxn],num[maxn];ll ma[maxn][maxn][maxn];int tot;int is_prime[maxn];ll mu[maxn];int prime[maxn];void Moblus(){    tot = 0;    mu[1] = 1;    memset(is_prime,0,sizeof(is_prime));    for(int i = 2; i < maxn-10; i++)    {        if(!is_prime[i])        {            prime[tot++] = i;            mu[i] = -1;        }        for(int j = 0; j < tot && i*prime[j] < maxn-10; j++)        {            is_prime[i*prime[j]] = 1;            if(i % prime[j])            {                mu[i*prime[j]] = -mu[i];            }            else            {                mu[i*prime[j]] = 0;                break;            }        }    }}ll dfs(int u,int gcd){    if(dp[u] != -1) return dp[u];    ll ans = 0;    for(int i = 1; i <= n; i++)    {        if(ma[gcd][u][i])        {            ans = (ll)(ans +  ma[gcd][u][i] + (ll)ma[gcd][u][i]*dfs(i,gcd)%mod)%mod;        }    }    return dp[u] = ans;}ll solve(int gcd){    clr(dp,-1);    ll ans = 0;    for(int i = 1; i <= n; i++)    {        if(dp[i] == -1)            dfs(i,gcd);    }    for(int i = 1; i <= n; i++)        ans = (ans + dp[i])%mod;    return ans;}void debug(){    for(int i = 1; i <= 10; i++)    {        num[i] = solve(i);        cout << num[i] <<" ";    }    cout << endl;}int u[maxn*maxn*5],vec[maxn*maxn*5];int v[maxn*maxn*5];int x[maxn*maxn*5];int main(){//    freopen("in.txt","r",stdin);    int y;    Moblus();    read(n),read(m);    memset(ma,0,sizeof(ma));    for(int i = 1; i <= m; i++)    {        read(u[i]),read(v[i]);        read(x[i]);        for(int j = 1; j * j <= x[i]; j++)        {            if(x[i] % j) continue;            ma[j][u[i]][v[i]] ++;            if(j * j != x[i])                ma[x[i]/j][u[i]][v[i]] ++;        }    }    for(int i = 1; i <= 100; i++)    {        num[i] = solve(i);    }//    debug();    read(q);    ll ans = 0;    for(int i = 1; i <= 100; i++)    {        ans = (ans + (ll)mu[i] * num[i] +mod)%mod;    }     printf("%I64d\n",ans);    int id,cnt;    for(int i = 1; i <= q; i++)    {        ans = 0,cnt = 0;        read(id),read(y);        int a = u[id],b = v[id];        for(int i = 1; i*i <= x[id]; i++)        {            if(x[id] % i) continue;            ma[i][a][b] --,vec[cnt++] = i;            if(i * i != x[id])                ma[x[id]/i][a][b] --,vec[cnt++] = x[id]/i;        }        x[id] = y;        for(int i = 1; i*i <= x[id]; i++)        {            if(x[id] % i) continue;            ma[i][a][b] ++,vec[cnt++] = i;            if(i*i != x[id])                ma[x[id]/i][a][b] ++,vec[cnt++] = x[id]/i;        }        for(int i = 0; i < cnt; i++)        {            num[vec[i]] = solve(vec[i]);        }//        debug();        for(int i = 1; i <= 100; i++)        {            ans = (ans + (ll)mu[i] * num[i]+mod)%mod;        }        printf("%I64d\n",ans);    }    return 0;}