Yuanfang is puzzled with the question below:
There are n integers, a 1, a 2, …, a n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a x and a y inclusive. In other words, do transformation a k<---a k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a x and a y inclusive. In other words, do transformation a k<---a k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a x and a y to c, inclusive. In other words, do transformation a k<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a x and a y inclusive. In other words, get the result of a x p+a x+1 p+…+a y p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.

Input

There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input

5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output

307
7489

解題思路：

硬來肯定是不行的，儘管人家給了8000ms。。剛開始初始值都是一樣的，經過一項操作之後該區間的數值也是一樣的，這裡就是突破點。查詢的時候肯定不能一個一個點算，但是如果某一區間的數值都相同的話，那該區間的和就是x^p * (r - l + 1)了（x為單點的數值，（r - l + 1）為區間長度，如此一來就能節約很多時間。當然，線段樹中的節點存放的就是該區間的單點的數值（如果該區間所有的數值都相同的話）。具體請看下面程式碼。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int mod = 10007;
const int maxn = 100020;
int n, m;
int Tree[maxn << 2];
bool flag[maxn << 2];

void Init()
{
	memset(flag, true, sizeof(flag));
	memset(Tree, 0, sizeof(Tree));
}

void update(int op, int L, int R, int val, int l, int r, int rt)
{
	if(l >= L && r <= R && flag[rt])
	{
		if(op == 1)
			Tree[rt] = (Tree[rt] + val) % mod;
		else if(op == 2)
			Tree[rt] = (Tree[rt] * val) % mod;
		else if(op == 3)
			Tree[rt] = val;
		return ;
	}
	if(flag[rt])
	{
		flag[rt << 1] = flag[rt << 1 | 1] = 1;
		flag[rt] = false;
		Tree[rt << 1] = Tree[rt << 1 | 1] = Tree[rt];
	}
	int m = (l + r) / 2;
	if(m >= L)
		update(op, L, R, val, l, m, rt << 1);
	if(m < R)
		update(op, L, R, val, m + 1, r, rt << 1 | 1);
	if(!flag[rt << 1] || !flag[rt << 1 | 1])
		flag[rt] = false;
	else
	{
		if(Tree[rt << 1] != Tree[rt << 1 | 1])
			flag[rt] = false;
		else 
		{
			flag[rt] = true;
			Tree[rt] = Tree[rt << 1];
		}
	}
}

int query(int L, int R, int p, int l, int r, int rt)
{
	if(l >= L && r <= R && flag[rt])
	{
		int ans = 1;
		for(int i = 1; i <= p; ++ i)
		{
			ans = (ans * Tree[rt]) % mod;
		}
		ans = ans * (r - l + 1) % mod;
		return ans;
	}
    if(flag[rt])
	{
		flag[rt<<1]=flag[rt<<1|1]=1; 
		flag[rt]=0; 
		Tree[rt<<1]=Tree[rt<<1|1]=Tree[rt]; 
	}
	int m = (l + r) / 2;
	int ans = 0;
	if(m >= L)
		ans += query(L, R, p, l, m, rt << 1);
	if(m < R)
		ans += query(L, R, p, m + 1, r, rt << 1 | 1);
	return ans % mod;
}

int main()
{
	while(cin >> n >> m)
	{
		if(n == 0 && m == 0)
			break;
		int x, y, op, num;
		Init();
		for(int i = 1; i <= m; ++ i)
		{
			scanf("%d%d%d%d", &op, &x, &y, &num);
			if(op <= 3)
			{
				update(op, x, y, num, 1, n, 1);
			}
			else
			{
				printf("%d\n", query(x, y, num, 1, n, 1));
			}
		}
	}
	return 0;
}