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 yp. 
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

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
const int mod=10007;
int tree[maxn*4],col[maxn*4];
int n,m,x,y,c,Q;

void pushup(int rt)
{
    if(!col[2*rt]||!col[2*rt+1]) col[rt]=0;//左右子樹有不同，那麼col為0代表有不同
    else if(tree[2*rt]!=tree[2*rt+1]) col[rt]=0;
    else col[rt]=1,tree[rt]=tree[2*rt];
}

void pushdown(int rt)
{
    if(col[rt])
    {
        col[2*rt]=col[2*rt+1]=1;
        tree[2*rt]=tree[2*rt+1]=tree[rt];
        col[rt]=0;
    }
}

void update(int l,int r,int rt)
{
    if(x<=l&&y>=r&&col[rt])
    {
        if(Q==1) tree[rt]=(tree[rt]+c)%mod;
        else if(Q==2) tree[rt]=(tree[rt]*c)%mod;
        else tree[rt]=c;
        return;
    }
    pushdown(rt);
    int mid=(l+r)/2;
    if(x<=mid) update(l,mid,2*rt);
    if(y>mid)  update(mid+1,r,2*rt+1);
    pushup(rt);
}

int query(int l,int r,int rt)
{
    if(x<=l&&y>=r&&col[rt])
    {
        ll ans=1;
        for(int i=1;i<=c;i++) ans=(ans*tree[rt])%mod;
        ans=(ans*(r-l+1))%mod;   //要是都一樣就可以直接乘了
        return ans;
    }
    pushdown(rt);
    int mid=(l+r)/2;
    int left=0,right=0;
    if(x<=mid) left+=query(l,mid,2*rt);
    if(y>mid)  right+=query(mid+1,r,2*rt+1);
    return (left+right)%mod;   //還要再mod一次，就wa在了這裡
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0) break;
        memset(tree,0,sizeof(tree));//最開始時col都為1，代表左右子樹裡的值最開始都是一樣的
        memset(col,1,sizeof(col));
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d%d",&Q,&x,&y,&c);
            if(Q>=1&&Q<=3)
            {
                update(1,n,1);
            }
            else
                printf("%d\n",query(1,n,1));
        }
    }
    return 0;
}