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：

線段樹的各種操作都是很簡單的，這道題的突破點就在於判斷區間的所有數是否相同，如果相同的話求他們的冪就很簡單了，也就是求出其中一個再乘上區間長度就好了。flag陣列判斷是否為相同的數，Tree陣列存放相同的數的數值

#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;
}