[线段树+欧拉函数+状态压缩] 区间修改 求区间积的欧拉函数 CF1114F

F. Please, another Queries on Array?

time limit per test

5.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array a1,a2,…,ana1,a2,…,an.

You need to perform qq queries of the following two types:

1. "MULTIPLY l r x" — for every ii (l≤i≤rl≤i≤r) multiply aiai by xx.
2. "TOTIENT l r" — print φ(∏i=lrai)φ(∏i=lrai) taken modulo 109+7109+7, where φφ denotes Euler's totient function.

The Euler's totient function of a positive integer nn (denoted as φ(n)φ(n)) is the number of integers xx (1≤x≤n1≤x≤n) such that gcd(n,x)=1gcd(n,x)=1.

Input

The first line contains two integers nn and qq (1≤n≤4⋅1051≤n≤4⋅105, 1≤q≤2⋅1051≤q≤2⋅105) — the number of elements in array aa and the number of queries.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤3001≤ai≤300) — the elements of array aa.

Then qq lines follow, describing queries in the format given in the statement.

1. "MULTIPLY l r x" (1≤l≤r≤n1≤l≤r≤n, 1≤x≤3001≤x≤300) — denotes a multiplication query.
2. "TOTIENT l r" (1≤l≤r≤n1≤l≤r≤n) — denotes a query on the value of Euler's totient function.

It is guaranteed that there is at least one "TOTIENT" query.

Output

For each "TOTIENT" query, print the answer to it.

Example

input

Copy

4 4
5 9 1 2
TOTIENT 3 3
TOTIENT 3 4
MULTIPLY 4 4 3
TOTIENT 4 4

output

Copy

1
1
2

Note

In the first example, φ(1)=1φ(1)=1 for the first query, φ(2)=1φ(2)=1 for the second query and φ(6)=2φ(6)=2 for the third one.

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int mn = 4e5 + 10;
const int mod = 1e9 + 7;

int pcnt, pm[100];
void shai()
{
	bool is[310];
    for (int i = 2; i <= 300; i++)
		is[i] = 1;
    for (int i = 2; i <= 300; i++)
	{
		if (!is[i])
			continue;
        pm[pcnt] = i;	// 300内从小到大质数表
        pcnt++;
        for (int j = 2; i * j <= 300; j++)
			is[i * j] = 0;
	}
}
ll yin[310], inv[310];
ll calzhi(int x)
{
	ll ans = 0;
    for (int i = 0; pm[i] <= x && i < pcnt; i++)
	{
		if (x % pm[i] == 0)	// x因子包含该质数
			ans += (1ll << i);
	}
	return ans;
}
ll ksm(ll x, ll y)
{
	ll ans = 1;
	while(y)
	{
		if (y & 1)
			ans = ans * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return ans;
}

struct node
{
	int l, r;
	ll ji, zhi, lazy, lazy2;
} tr[8 * mn];

void pushup(int id)
{
	tr[id].ji = tr[2 * id].ji * tr[2 * id + 1].ji % mod;
    tr[id].zhi = tr[2 * id].zhi | tr[2 * id + 1].zhi;
}
void pushdown(int id)
{
    if (tr[id].lazy == 1 || tr[id].lazy2 == 0)
		return;
	
    tr[2 * id].lazy = tr[2 * id].lazy * tr[id].lazy % mod;
    tr[2 * id + 1].lazy = tr[2 * id + 1].lazy * tr[id].lazy % mod;
    tr[2 * id].ji = tr[2 * id].ji * ksm(tr[id].lazy, tr[2 * id].r - tr[2 * id].l + 1) % mod;
    tr[2 * id + 1].ji = tr[2 * id + 1].ji * ksm(tr[id].lazy, tr[2 * id + 1].r - tr[2 * id + 1].l + 1) % mod;
    tr[id].lazy = 1;
    
    tr[2 * id].lazy2 |= tr[id].lazy2;
    tr[2 * id + 1].lazy2 |= tr[id].lazy2;
    tr[2 * id].zhi |= tr[2 * id].lazy2;
    tr[2 * id + 1].zhi |= tr[2 * id + 1].lazy2;
	tr[id].lazy2 = 0;
}

void build(int l, int r, int id)
{
    tr[id].l = l, tr[id].r = r, tr[id].lazy = 1, tr[id].lazy2 = 0;
    if (l == r)
	{
        int t;
        scanf("%d", &t);
        tr[id].ji = t;
        tr[id].zhi = yin[t];
        return;
	}
    int mid = (l + r) >> 1;
    build(l, mid, 2 * id);
    build(mid + 1, r, 2 * id + 1);
    pushup(id);
}
void change(int id, int L, int R, int x, ll xzhi)
{
    int l = tr[id].l, r = tr[id].r;
    if (L <= l && r <= R)
	{
        tr[id].ji = tr[id].ji * ksm(x, tr[id].r - tr[id].l + 1) % mod;
        tr[id].zhi = tr[id].zhi | xzhi;
        tr[id].lazy = tr[id].lazy * x % mod;
        tr[id].lazy2 |= xzhi;
		return;
	}
	pushdown(id);
    int mid = (l + r) >> 1;
    if (R <= mid)
		change(2 * id, L, R, x, xzhi);
	else if (mid +1 <= L)
		change(2 * id + 1, L, R, x, xzhi);
    else 
    {
    	change(2 * id, L, mid, x, xzhi);
    	change(2 * id + 1, mid + 1, R, x, xzhi);
    }
	
	pushup(id);
}
ll anszhi, ansji;
void query(int id, int L, int R)
{
	int l = tr[id].l, r = tr[id].r;
	if (L <= l && r <= R)
	{
		ansji = ansji * tr[id].ji % mod;
        anszhi |= tr[id].zhi;
		return;
	}
	pushdown(id);
    int mid = (l + r) >> 1;
    if (R <= mid)
		query(2 * id, L, R);
	else if (mid + 1 <= L)
		query(2 * id + 1, L, R);
	else 
	{
		query(2 * id, L, mid);
		query(2 * id + 1, mid + 1, R);
	}
}

int main()
{
	shai();
	for (int i = 1; i <= 300; i++)
	{
		yin[i] = calzhi(i);	// i 包含的质数 
        // 300以内素数62个,是否含有 = 0/1, 压缩进一个long long
		inv[i] = ksm(i, mod - 2);	// 求 i 的逆元
	}

	int n, q;
	scanf("%d %d", &n, &q);
	build(1, n, 1);

	char ch[50];
	int l, r, x;
	while (q--)
	{
		scanf("%s", ch);
		if (ch[0] == 'M')
		{
			scanf("%d %d %d", &l, &r, &x);
			change(1, l, r, x, yin[x]);
			// 区间 * x; 区间包含的质数 | x包含的质数
		}
		else if (ch[0] == 'T')
		{
			anszhi = 0, ansji = 1;
			scanf("%d %d", &l, &r);
            query(1, l, r);
            
            ll ans = ansji;
            for (int i = 0; i < pcnt; i++)
			{
				if ((anszhi >> i) & 1)
				ans = ans * (pm[i] - 1) % mod * inv[pm[i]] % mod;
				/// n的欧拉函数值 = n * (1 - 1 / p1) * (1 - 1 / p2) * ... * (1 - 1 / pn) 
				/// 1 - 1 / p1  ==  (p1 - 1) / p1  ==  (p1 - 1) * inv[p1]
			}
			printf("%lld\n", ans);
		}
	}
    return 0;
}