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;
using namespace std;
const int maxn = 100005;
struct node {
  long long sum, del;
  int x;
} s[maxn << 2];
void updata(int root, int l, int r) {
  s[root].sum =
      s[root << 1].sum + s[root << 1 | 1].sum + 1LL * (r - l + 1) * s[root].del;
}
void build(int root, int l, int r) {
  s[root].sum = s[root].del = 0;
  s[root].x = 0;
  if (l == r) {
    s[root].sum = 0;
    s[root].x = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(root << 1, l, mid);
  build(root << 1 | 1, mid + 1, r);
}
void gao(int root, int l, int r, int x) {
  if (s[root].x) {
    s[root].del += abs(s[root].x - x);
    s[root].sum += 1LL * (r - l + 1) * abs(s[root].x - x);
  } else {
    int mid = (l + r) >> 1;
    gao(root << 1, l, mid, x);
    gao(root << 1 | 1, mid + 1, r, x);
    updata(root, l, r);
  }
}
void paint(int l, int r, int root, int ll, int rr, int x) {
  if (l > rr || r < ll) return;
  if (ll <= l && r <= rr) {
    gao(root, l, r, x);
    s[root].x = x;
    return;
  }
  if (s[root].x) {
    s[root << 1].x = s[root].x;
    s[root << 1 | 1].x = s[root].x;
    s[root].x = 0;
  }
  int mid = (l + r) >> 1;
  paint(l, mid, root << 1, ll, rr, x);
  paint(mid + 1, r, root << 1 | 1, ll, rr, x);
  updata(root, l, r);
}
long long query(int l, int r, int root, int ll, int rr) {
  if (l > rr || r < ll) return 0;
  if (ll <= l && rr >= r) return s[root].sum;
  int mid = (l + r) >> 1;
  return query(l, mid, root << 1, ll, rr) +
         query(mid + 1, r, root << 1 | 1, ll, rr) +
         s[root].del * (min(rr, r) - max(ll, l) + 1);
}
int n, m;
int main() {
  int l, r, x, typ;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%d", &typ);
    scanf("%d%d", &l, &r);
    if (typ == 1) {
      scanf("%d", &x);
      paint(1, n, 1, l, r, x);
    } else {
      printf("%I64d\n", query(1, n, 1, l, r));
    }
  }
  return 0;
}
