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 <bits/stdc++.h>
using namespace std;
struct TT {
  long long int a, b;
  TT(long long int x = 0, long long int y = 0) {
    a = x;
    b = y;
  }
  TT operator+(const struct TT &m) const { return TT(a + m.a, b + m.b); }
} tree[1 << 18];
struct node {
  int l, r;
  long long int x;
  node(int a = 0, int b = 0, long long int c = 0) {
    l = a;
    r = b;
    x = c;
  }
  bool operator<(const struct node &m) const { return l < m.l; }
};
set<node> myset;
int n, m;
void init() {
  myset.insert(node(0, 0, 0));
  myset.insert(node(n + 2, 0, 0));
  for (int i = 1; i <= n; i++) {
    myset.insert(node(i, i, i));
  }
}
void print() {
  for (set<node>::iterator it = myset.begin(); it != myset.end(); it++) {
    printf("%d %d %lld\n", (*it).l, (*it).r, (*it).x);
  }
}
void updateT(int ind, long long int v) {
  TT add = TT((long long int)(-ind + 1) * v, (long long int)v);
  while (ind <= 100008) {
    tree[ind] = tree[ind] + add;
    ind += (ind & -ind);
  }
}
void update(int l, int r, long long int x) {
  updateT(l, x);
  updateT(r + 1, -x);
}
TT read(int ind) {
  TT sum = 0;
  while (ind != 0) {
    sum = sum + tree[ind];
    ind = ind - (ind & -ind);
  }
  return sum;
}
void Update(int l, int r, long long int x) {
  node a[2];
  int yes[2] = {0, 0};
  ;
  int counter = 0;
  set<node>::iterator f1, f2, arr[2];
  f1 = myset.lower_bound(l);
  f2 = myset.lower_bound(r);
  f1--;
  f2++;
  int e1, e2;
  for (set<node>::iterator it = f1; it != f2; it++) {
    if (!((*it).r < l || (*it).l > r)) {
      arr[counter] = it;
      if (counter == 0) counter = 1;
      arr[counter] = it;
      if ((*it).l >= l && (*it).l <= r)
        e1 = (*it).l;
      else {
        e1 = l;
      }
      if ((*it).r >= l && (*it).r <= r)
        e2 = (*it).r;
      else {
        e2 = r;
      }
      if ((*it).l < l) {
        yes[0] = 1;
        a[0] = node((*it).l, l - 1, (*it).x);
      }
      if ((*it).r > r) {
        yes[1] = 1;
        a[1] = node(r + 1, (*it).r, (*it).x);
      }
      update(e1, e2, abs((*it).x - x));
    }
  }
  arr[1]++;
  myset.erase(arr[0], arr[1]);
  myset.insert(node(l, r, x));
  if (yes[0]) myset.insert(a[0]);
  if (yes[1]) myset.insert(a[1]);
}
long long int getsum(TT a, int i) { return a.a + (long long int)i * a.b; }
void solve() {
  int type, l, r;
  scanf("%d%d", &n, &m);
  init();
  long long int x;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%lld", &l, &r, &x);
      Update(l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", getsum(read(r), r) - getsum(read(l - 1), l - 1));
    }
  }
}
int main() {
  solve();
  return 0;
}
