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 node {
  int lv, rv;
  long long val, prom;
  node *l, *r;
  node() {
    l = r = NULL;
    val = prom = 0;
    lv = 1e9;
    rv = -1e9;
  }
  long long len() { return (long long)rv - lv + 1; }
} * root;
void build(node *v) {
  if (v->lv == v->rv) return;
  int m = (v->lv + v->rv) / 2;
  node *l = new node, *r = new node;
  l->lv = v->lv;
  l->rv = m;
  v->l = l;
  r->lv = m + 1;
  r->rv = v->rv;
  v->r = r;
  build(l);
  build(r);
}
void push(node *v) {
  if (v->l) {
    (v->l)->val += v->prom * (v->l)->len();
    (v->l)->prom += v->prom;
  }
  if (v->r) {
    (v->r)->val += v->prom * (v->r)->len();
    (v->r)->prom += v->prom;
  }
  v->prom = 0;
}
void add_range(node *v, int l, int r, long long dif) {
  if (v->lv > r || v->rv < l) return;
  if (l <= v->lv && v->rv <= r) {
    v->val += v->len() * dif;
    v->prom += dif;
    return;
  }
  push(v);
  v->val = 0;
  if (v->l) {
    add_range(v->l, l, r, dif);
    v->val += v->l->val;
  }
  if (v->r) {
    add_range(v->r, l, r, dif);
    v->val += v->r->val;
  }
}
long long get(node *v, int l, int r) {
  if (v->lv > r || v->rv < l) return 0;
  if (l <= v->lv && v->rv <= r) return v->val;
  push(v);
  long long res = 0;
  if (v->l) res += get(v->l, l, r);
  if (v->r) res += get(v->r, l, r);
  return res;
}
struct tt {
  int l, r, col;
  tt() {}
  tt(int l, int r, int col) : l(l), r(r), col(col) {}
};
bool operator<(tt a, tt b) { return a.l < b.l; }
set<tt> s;
void colorize() {
  int l, r, col;
  scanf("%d%d%d", &l, &r, &col);
  l--;
  set<tt>::iterator be, en;
  vector<tt> to_add, to_erase;
  to_add.push_back(tt(l, r, col));
  be = s.lower_bound(tt(l, -1, -1));
  en = s.lower_bound(tt(r, -1, -1));
  if (be != s.begin()) be--;
  if (en != s.end()) en++;
  if (en != s.end()) en++;
  for (; be != en; be++) {
    if (be->l >= r || be->r <= l) continue;
    if (be->l < l) to_add.push_back(tt(be->l, l, be->col));
    if (be->r > r) to_add.push_back(tt(r, be->r, be->col));
    to_erase.push_back(*be);
    add_range(root, max(be->l, l), min(be->r, r) - 1, abs(col - be->col));
  }
  for (tt i : to_erase) s.erase(s.find(i));
  for (tt i : to_add) s.insert(i);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  root = new node;
  root->lv = 0;
  root->rv = n - 1;
  build(root);
  for (int i = 0; i < n; i++) s.insert(tt(i, i + 1, i + 1));
  while (m--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      colorize();
      continue;
    }
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    printf("%lld\n", get(root, l, r));
  }
  exit(0);
}
