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 l, r, color;
  long long sum, add;
  node *lch, *rch;
} * root, pool[(101000) << 1];
int cnt;
void doadd(node *x, long long v) {
  x->add += v;
  x->sum += (x->r - x->l + 1) * v;
}
void pushdown(node *x) {
  if (x->lch == nullptr) return;
  if (x->add) {
    doadd(x->lch, x->add), doadd(x->rch, x->add);
    x->add = 0;
  }
  if (~x->color) {
    x->lch->color = x->color;
    x->rch->color = x->color;
  }
}
void pushup(node *x) {
  x->sum = x->lch->sum + x->rch->sum;
  if (x->lch->color == x->rch->color)
    x->color = x->lch->color;
  else
    x->color = -1;
}
node *build(int l, int r) {
  node *x = &pool[cnt++];
  x->l = l, x->r = r;
  x->sum = x->add = 0;
  if (l == r) {
    x->lch = x->rch = nullptr;
    x->color = l;
  } else {
    int mid = (l + r) >> 1;
    x->lch = build(l, mid);
    x->rch = build(mid + 1, r);
    pushup(x);
  }
  return x;
}
void change(node *x, int l, int r, int c) {
  if ((~x->color) && l <= x->l && x->r <= r) {
    doadd(x, abs(x->color - c));
    x->color = c;
    return;
  }
  int mid = (x->l + x->r) >> 1;
  pushdown(x);
  if (l <= mid) change(x->lch, l, r, c);
  if (r > mid) change(x->rch, l, r, c);
  return pushup(x);
}
long long sum(node *x, int l, int r) {
  if (l <= x->l && x->r <= r) return x->sum;
  int mid = (x->l + x->r) >> 1;
  pushdown(x);
  if (l <= mid && mid < r) return sum(x->lch, l, r) + sum(x->rch, l, r);
  if (l <= mid) return sum(x->lch, l, r);
  return sum(x->rch, l, r);
}
int main() {
  int n, m;
  cin >> n >> m;
  root = build(1, n);
  for (int i = 1, op, l, r, x; i <= m; ++i) {
    cin >> op >> l >> r;
    if (op == 1) {
      cin >> x;
      change(root, l, r, x);
    } else
      cout << sum(root, l, r) << endl;
  }
  return 0;
}
