CodeForces

F - DZY Loves Colors

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

1. Paint all the units with numbers between l and r (both inclusive) with color x.
2. Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 10⁵).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 10⁸) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3

Output

8

Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3

Output

3
2
1

Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10

Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

#include <cstdio>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>

#define lid id<<1
#define rid id<<1|1
#define closein cin.tie(0)
#define scac(a) scanf("%c",&a)
#define scad(a) scanf("%d",&a)
//#define print(a) printf("%d
",a)
#define debug printf("hello world")
#define form(i,n,m) for(int i=n;i<m;i++)
#define mfor(i,n,m) for(int i=n;i>m;i--)
#define nfor(i,n,m) for(int i=n;i>=m;i--)
#define forn(i,n,m) for(int i=n;i<=m;i++)
#define scadd(a,b) scanf("%d%d",&a,&b)
#define memset0(a) memset(a,0,sizeof(a))
#define scaddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scadddd(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)

#define INF 0x3f3f3f3f
#define maxn 100005
typedef long long ll;
using namespace std;
//---------AC(^-^)AC---------\

int n,m,block,num;
int blo[maxn],l[maxn],r[maxn];
ll color[maxn],sum[maxn],allsum[maxn],flag2[maxn],flag[maxn];

void print()
{
    forn(i,1,num)
    {
        printf("%lld %lld
",allsum[i],flag2[i]);
    }
    debug;
    printf("
");
    forn(i,1,n)
        printf("%lld %lld
",color[i],sum[i]);
}
void push_down(int id)
{
    forn(i,l[id],r[id]) color[i]=flag[id];
    flag[id]=-1;
}
void update(int ld,int rd,int add)
{
    if(blo[ld]==blo[rd])
    {
        if(flag[blo[ld]]!=-1)    push_down(blo[ld]);
        forn(i,ld,rd)    {
            sum[i]+=abs(color[i]-add);
            allsum[blo[ld]]+=abs(color[i]-add);
            color[i]=add;
        }
    }else {
        if(flag[blo[ld]]!=-1) push_down(blo[ld]);
        forn(i,ld,r[blo[ld]]) {
            sum[i]+=abs(color[i]-add);
            allsum[blo[ld]]+=abs(color[i]-add);
            color[i]=add;
        }
        forn(i,blo[ld]+1,blo[rd]-1)    {
            if(flag[i]!=-1)    {
                allsum[i]+=abs(flag[i]-add)*(r[i]-l[i]+1);
                flag2[i]+=abs(flag[i]-add);
                flag[i]=add;
            }else {
                forn(j,l[i],r[i])    {
                    sum[j]+=abs(color[j]-add);
                    allsum[i]+=abs(color[j]-add);
                    color[j]=add;
                }
                flag[i]=add;
            }
        }
        if(flag[blo[rd]]!=-1) push_down(blo[rd]);
        forn(i,l[blo[rd]],rd)    {
            sum[i]+=abs(color[i]-add);
            allsum[blo[rd]]+=abs(color[i]-add);
            color[i]=add;
        }
    }
    //print();
}
void query(int ld,int rd)
{
    ll ans=0;
    if(blo[ld]==blo[rd])    {
        forn(i,ld,rd) ans+=sum[i]+flag2[blo[i]];
    }else {
        forn(i,ld,r[blo[ld]])    {
            ans+=sum[i]+flag2[blo[i]];
        }
        forn(i,blo[ld]+1,blo[rd]-1)    {
            ans+=allsum[i];
        }
        forn(i,l[blo[rd]],rd)    {
            ans+=sum[i]+flag2[blo[i]];
        }
    }
    printf("%lld
",ans);
}

int main()
{
    scadd(n,m);
    block=sqrt(n);
    num=(n-1)/block+1;
    forn(i,1,n)    {
        color[i]=i;
        blo[i]=(i-1)/block+1;
    }
    forn(i,1,num) {
        l[i]=(i-1)*block+1;
        r[i]=i*block;
    }
    r[num]=n;
    forn(i,1,num) flag[i]=-1;
    while(m--)
    {
        int op,l,r;
        scaddd(op,l,r);
        if(op==1)
        {
            int x;
            scad(x);
            update(l,r,x);
        }
        else
        {
            query(l,r);
            //print();
        }
    }
    return 0;
}